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21. M E T H O D S^ O F
D I F F E R E N T I A T I O N
A N D A P P L I C A T I O N S
O F D E R I V A T I V E S
1. INTRODUCTION
The rate of change of one dependent quantity with respect to another dependent quantity has great importance. E.g. the rate of change of displacement of a particle with respect to time is called its velocity and the rate of change of velocity is called its acceleration. The rate of change of a quantity ‘y’ with respect to another quantity ‘x’ is known as the derivative or differentiable coefficient of ‘y’ with respect to ‘x.’
According to the first principle of calculus, if y = f(x) is the derivative function, then the derivative of f(x) with respect to x is given by:
f’(x) = dy dx
h 0 lim →
f(x+h)-f(x) h
Note: y’, y 1 , Dy can also be used to denote the derivative of y with respect to x. Differentiation is the process of
finding the derivative of a function. ⇒ sin β > 0; cos α < 0
2. DERIVATIVES OF SOME STANDARD FUNCTIONS
Different types of differentiation formulae
(a) d dx
(constant) = 0 (f) d dx
(xn) = nxn–
(b) d dx (ex) = ex^ (g) d dx (ax) = ax^ loge a
(c) d dx
(loge x) = 1 x
(h) d dx
(logax) = e
xlog a
(d) d dx
(sin x) = cos x (i) d dx
(cos x) = – sin x
(e) d dx
(tan x) = sec^2 x (j) d dx
(cot x) = – cosec^2 x
21.2 | Methods of Differentiation and Applications of Derivatives
(k) d dx
(sec x) = sec x tan x (z) d dx
(cosec x) = – cosec x cot x
(l) d dx
(sin–1x) = 2
1 −x
, – 1 < x < 1 (aa) d dx
(cos–1x) = – 2
1 −x
,– 1 < x < 1
(m) d dx (tan–1x) = (^2)
1 + x
, x ∈ R (ab) d dx (cot–1x) = – (^2)
1 + x
, ∀ x ∈ R
(n) d dx
(sec–1x) = 2
| x | x − 1
| x | > 1 (ac) d dx
(cosec–1x) = 2
| x | x 1
| x | >
(o) d dx
(sinh x) = coshx (ad) d dx
(coshx) = sinhx
(p) d dx
(tanh x) = sech^2 x (ae) d dx
(cothx) = –cosech^2 x
(q) d dx
(sechx) = – sechx tanhx (af) d dx
(cosechx) = –cosechx cothx
(r) d dx
(sinh–1x) = 2
1 +x
, ∀ x ∈ R (ag) d dx
(cosh–1x) = 2
x − 1
, | x | > 1
(s) d dx
(tanh–1x) = (^12) 1 − x
, x ± 1 (ah) d dx
(coth–1x) = 21 x − 1
, x ≠ ± 1
(t) d dx
(sech–1x) = – 2
| x | 1 −x
, | x | < 1 (ai) d dx
(cosech–1x) = 2
| x | x 1
,∀ x ∈ R
(u) d dx
(eax^ sin bx) = eax^ (a sin bx + b cos bx) = a^2 + b^2 eaxsin (bx + tan–1^ b/a)
(v) d dx
(eax^ cos bx) = eax^ (a cos bx – b sin bx) = a^2 + b^2 eaxcos (bx + tan–1^ b/a)
(w) d dx
| x | = x | x |
(x ≠ 0) (aj) d dx
log | x | = 1 x
, (x ≠ 0)
(x) d dx [x] = 0, ∀ x ∈ –I (where [. ] denotes greatest integer function)
(y) d dx
{x} = 1, ∀ x ∈ R (where {. } denotes fractional part function)
If the function is continuous, you do not have to apply the first principle method to check differentiability. You can go directly for dy/dx and check whether dy/dx exists on both the left and right sides and are equal. If dy/dx does not exist for either one side or both the sides or if both the derivatives exist, but are not equal or finite, then the function is not differentiable. E.g. Let y = sin(x) be a continuous function. Check differentiability at x = π/2.On checking for dy/dx = cos(x) on both the right and left sides, it is found to be equal and finite. Hence, y = sin(x) is differentiable at x = π/2.
MASTERJEE CONCEPTS
21.4 | Methods of Differentiation and Applications of Derivatives
Illustration 3: If y =
–1 1 1 1
tan x cot x tan x cot x
− − −
, find x 1
dy dx (^) =
. (JEE MAIN)
Sol: Differentiate and put x = 1.
y = (^) ( –1^1 )^1
tan x cot x ...... tan cot x 2 − −^ ^ −^ + − =π π ^
⇒ dy dx
π(1 + x )
π(1 + x )
π(1 + x )
x 1
dy dx (^) =
2 × π
π
Illustration 4: Differentiate the following functions with respect to x: (JEE MAIN)
(i) 2
3x 2 2x 4
(ii) sec^2 x e + 3cos–1x (iii) log 7 (log x)
Sol: (i) Let y = 2 3x 2 1 2x 4
= (3x + 2)1/2^ + (2x^2 + 4)–1/
dy dx
(^1 ) (^1) (3x 2) 2 2
−
(3x + 2) +
(^1 ) (2x^2 4)^2
− (^) −
(2x^2 + 4)
1 (^1) (3x 2) 2 2
− +. (3) –
3 (2x 2 + 4)−^2. 4x = 3 2 3x + 2
(ii) Let y = sec^2 x e + 3 cos–1^ x
dy dx
sec^2 x e. d dx
(sec^2 x) + 3 2
1 x
sec^2 x e. 2sec x d(sec x) dx
2
1 x
= 2 sec x (sec x tan x) sec^2 x e + 3 2
1 x
= 2 sec^2 x tan x sec^2 x e – 3 2
1 x
(iii) Let y = log 7 (log x) = log(logx) log
(using change of base formula)
dy dx
1 d log7 dx (log(logx)) =
log
logx
d dx
(log x) =
xlog7logx
Illustration 5: Find dy dx
, if y =3 tan x + 5 logax + x – 3ex^ + 1 x
. (JEE MAIN)
Sol: Chain rule.
We have y = 3 tan x + 5 logax + x – 3ex^ + 1 x
Thus, dy dx
= d dx
x a 3tanx 5log x x 3e 1 x
+^ +^ −^ +
= d dx
(3 tan x) + d dx
(5 loga x) + d dx
( x ) – d dx
(3ex) + d dx
x
= 3 sec^2 x + 5 x
(logea)-1+ 1 2
1 x−^2 – 3ex^ – x–^.
Mathematics | 21.
Illustration 6: Let f, g and h be differentiable functions. If f(0) = 1, g(0) = 2, h(0) = 3 and the derivative pairwise products at x = 0 are (fg)’ (0) = 6, (gh)’(0) =4 and (hf)’ (0) = 5, then compute the value of (fgh)’ (0). (JEE ADVANCED)
Sol: Product rule
(fgh)’ = f’gh + fhg’ + fgh’ …….(i)
(fg)’ (0) = 6 ⇒ (fg’ + gf’) (0) = 6
(gh)’ (0) = 4 ⇒ (gh’ + hg’) (0) = 4
(hf)’ (0) = 5 ⇒ (hf’ + fh’) (0) = 5
(fgh)’ = 1 2
(2f’gh + 2fg’h + 2fgh’) = 1 2
(f’gh + f’gh + fg’h + fg’h + fgh’ + fgh’)
[h(f’g + fg’) + g(f’h + fh’) + f(g’h + gh’)] =
[h(fg)’ + g(fh)’ + f(gh)’]
⇒ (fgh)’ (0) = 1 2
[(3)(6) + (2)(5) + (1)(4)]= 1
[18 + 10 + 4] = 16
6. TRIGONOMETRIC TRANSFORMATIONS
In case of inverse trigonometric functions, it becomes very easy to differentiate a function by using trigonometric transformations.Given below are some important results on trigonometric and inverse trigonometric functions.
(a) sin2x = 2 sin x cos x = 2tanx 2 1 +tan x
(b) cos 2x = 2 cos^2 x –1 = 1 – 2 sin^2 x = cos^2 x – sin 2 x =
2 2
1 tan x 1 tan x
(c) tan 2x = 2tanx 2 1 – tan x
(n) sin 3x = 3 sin x – 4 sin 3 x
(d) cos 3x = 4 cos 3 x – 3 cos x (o) tan 3x =
3 2
3tanx tan x 1 – 3tan x
(e) sin –1^ x + cos–1^ x = tan–1^ x + cot–1^ x =sec–1^ x + cosec–1^ x = 2
π
(f) sin –1^ (–x) = – sin–1^ x (p) cos –1^ (–x) = π – cos –1^ x
(d) tan–1^ (–x) = – tan–1^ x (q) cosec –1^ (–x) = – cosec–1^ x
(h) cot–1^ (–x) = π – cot–1^ x (r) sec –1^ (–x) = π – sec –1^ x
(i) sin –1^ x ± sin–1^ y = sin–^ (x 1 − y^2 ± y 1 − x^2 ) (s) cos –1^ x ± cos–1^ y = cos–^ (xy 1 − x 2 1 −y 2 )
(j) tan–1^ x ± tan–1^ y = tan–
x y 1 xy
(t)^ 2sin^
–1 (^) x = sin–1 (^) (2x 1 − x^2 ) (Be aware of ranges for ‘x’)
(k) 2cos –1^ x = cos–1^ (2x^2 – 1) (u) 2 tan–1x = tan–1^2
2x 1 x
− = sin
–1 2x 2 1 x
+ =cos
2 2
1 x 1 x
(l) 4
π (^) – tan–1x = tan–1^1 x 1 x
(v) 3 sin–1x = sin–1^ (3x – 4x^3 )
(m) 3 cos–1x = cos–1^ (4x^3 – 3x) (w) 3 tan–1x = tan–
3 2
3x x 1 3x
Mathematics | 21.
To differentiate a complex function, put x in some trigonometric form so that the function can be easily differentiated and then put back x in the form of an inverse trigonometric function. E.g. Find the derivatives of sec –1^ [1/(2x^2 – 1)] with respect to 1 − x^2 at x = 1/2. Sol. Putting x = cosθ, we get
u = sec–1^2
2cos θ − 1 = sec
–1(sec2θ) = 2θ and y = 1 − x^2 = sinθ
\ u = 2sin–1^ y ⇒ du dy
2
1 −y
2
x
Thus, x 1/
du (^4) dy (^) =
Ravi Vooda (JEE 2009, AIR 71)
Illustration 8: If y = (a − x)(x − b) – (a – b) tan–
a x x b
, find dy dx
. (JEE ADVANCED)
Sol: Use Substitution to simplify the given expression and then differentiate.
Let x = a cos^2 θ + b sin^2 θ
\ a – x = a – a cos^2 θ – b sin 2 θ = (a– b) sin^2 θ … (i)
x – b = a cos^2 θ + b sin^2 θ – b = (a – b) cos^2 θ … (ii)
\ y = (a – b) sinθ cosθ – (a – b) tan–1^ (tanθ)
y = (a^ b) 2
− (^) sin2θ – (a – b) θ
Then, dy dx
dy d dx d
(^) θ (^) θ
= (a^ b)cos2^ (a^ b) (b a)sin
− θ − − − θ
= 1 cos sin
− θ θ
= tan θ= a^ x x b
[From (i) and (ii)]
7. LOGARITHMIC DIFFERENTIATION
If differentiation of an expression is done after taking log on both the sides, then it is known as logarithmic differentiation. This method is used when a given expression is in one of the following forms:
(a) (f(x))g(x)
Let y = (f(x))g(x)
Taking logarithm of both the sides, we get log y = g(x) log f(x)
Differentiating with respect to x, we get
1 y
. dy dx
= g(x). 1 f(x)
. f’(x) + log f(x). g’(x) ⇒ dy dx
= y g(x) f '(x) logf(x).g'(x) f(x)
⇒ dy dx
= (f(x))g(x)
g(x) f '(x) logf(x).g'(x) f(x)
MASTERJEE CONCEPTS
21.8 | Methods of Differentiation and Applications of Derivatives
Short method: The derivative of [f(x)]g(x)can be directly written as:
d (^) (f(x)) g(x) f(x)g(x) d{g(x)logf(x)} dx dx
(b) Product of three or more functions
If y = f(x).g(x).h(x), then y’ = f(x).g(x).h(x). f '(x)^ g'(x) h'(x) f(x) g(x) h(x)
+^ +
Illustration 9: If f(x) =
(^100) n(101 n) n 1
(x n) −
∏ − , find f’(101)/f(101).^ (JEE ADVANCED)
Sol: Use logarithms followed by differentiation.
f(x) =
100 n(101 n) n 1
(x n) −
∏ −
ln f(x) =
100 n 1
n(101 n)ln(x n)
∑ −^ − ⇒
100 n 1
f '(x) n(101 n) f(x) (^) = x n
∑ −
⇒ f '(101) f(101)
100 n 1
n(101 n) 100x =^101 n^2
∑ (^) −
Illustration 10: Find the derivative of (sin x)cosx^. (JEE MAIN)
Sol: Take logarithms on both sides and differentiate.
d dx
(sin x)cos x^ = (sin x)cos x
d {cos xlog(sinx)} dx
= (sin x)cos x^ [cos x. cot x – sin x. log sin x]
Illustration 11: Find the derivative of xx^ with respect to x. (JEE MAIN)
Sol: Use logarithms to find the derivative.
Let y = xx^ or y = xx
log y = x log x = ex ln x
1 y
dy dx
= x 1 x
= d dx
exlnx^ = ex ln x
d dx (x ln x)
dy dx
= xx(1 + log x) (or) = ex ln x^ x. 1 lnx x
\ d x x^ xx (^) ( 1 log xe ) dx = + (or)^ = xx(1 + ln x)
Hence d dx
(xx) = xx(1 + ln x)
Illustration 12: Differentiate xsin x with respect to x. (JEE MAIN)
Sol: Similar to the previous illustration.
First method: Let y = xsin x
21.10 | Methods of Differentiation and Applications of Derivatives
dy dx
cos 1 x.log x e
− (^) d dx
(cos–1x. log x) ⇒ dy dx
cos 1 x d^1 1 d x logx. (cos x) cos x. (logx) dx dx
⇒ dy dx
cos 1 x^1 2 x logx^ cos^ x 1 x x
− ^ −^ −
(ii) Let cos 1 x (sinx) −
Then, y = cos 1 x.log sin x e −
Differentiating both the sides with respect to x, we get
dy dx
cos 1 x.logsin x e − (^) d dx
(cos–1x. log sin x)
dy dx
cos 1 x 1 d^ d – (sinx) cos x. (log sin x) log sinx (cos x) dx dx
⇒ dy dx
cos 1 1 2
(sinx) x cos. 1 cos x log sin x^1 sinx (^1) x
− ^ − ^ −
+ ^
^ −
⇒ dy dx
cos 1 x 1 2
(sinx) cos x. cot x logsinx 1 x
− ^ −
Illustration 17: Find dy dx
, if y =
sin x^ ..... (sinx)sin x
∞ (JEE ADVANCED)
Sol: Write the given expression as y = (sin x)y^ and proceed.
We have y = (sin x)y^ , Therefore, log y = y log sin x
Differentiating both the sides with respect to x, we get
1 y
dy dx
= y d dx
(log sin x) + (log sin x) dy dx
= y cos x sinx
⇒ 1 logsinx y
dy dx
= y cot x or dy dx
y 2 cot x 1 − y logsinx
y^2 cot x 1 −log y
8. DIFFERENTIATION OF IMPLICIT FUNCTION
If in an equation, both x and y occur together, i.e. f(x, y) = 0, and the equation cannot be solved for either x or y, then x (or y) is called the implicit function of y (or x).
E.g. x^3 + y^3 + 3axy + c = 0, xy^ + yx^ = ab, etc.
Working rule for finding the derivative
First method:
(a) Every term of f(x, y) = 0 should be differentiated with respect to x.
(b) The value of dy/dx should be obtained by rearranging the terms.
Second method:
If f(x, y) = constant, then dy dx
f / x f / y
, where f x
and f y
are the partial differential coefficients of f(x, y) with
respect to x and y, respectively.
Mathematics | 21.
Note: Partial differential coefficient of f(x, y) with respect to x can be defined as the ordinary differential coefficient of f(x, y) with respect to x keeping y constant.
E.g. z x y^2 z^ x ,^2 x 2xy y x
= ⇒ ∂^ = ∂ =
Illustration 18: If y = sinx 1 cos x 1 sinx 1 cos x 1 ....to
, then prove that dy dx
= (1^ y)cos x^ y sinx 1 2y cos x sinx
. (JEE ADVANCED)
Sol: Write the R.H.S. in terms of x and y. Then differentiate the equation on both sides.
We have, y = ( )
sinx 1 + (cos x) / (1 + y)
(1 y)sinx 1 y cos x
⇒ y + y^2 + y cos x = (1 + y) sin x
On differentiating both the sides with respect to x, we get
dy dx
dy dx
cos x – y sin x = dy dx
sin x + (1 + y) cos x
⇒ dy dx
{1 + 2y + cos x – sin x} = (1 + y) cos x + y sin x ⇒ dy dx
(1 y)cos x y sinx 1 2y cos x sinx
Illustration 19: If f(x) = (^) x 1 2x 1 2x 1 2x ......
, then compute the value of f (100). f’(100). (JEE MAIN)
Sol: Same as above y – x = 1 2x 1 2x 1 2x ......
⇒ y – x = 1 2x + y − x
⇒ (y – x) (x + y ) = 1 ⇒ y^2 – x^2 = 1
⇒ (f(x))^2 = 1 + x^2 ⇒ 2(f(x)) × f’(x) = 2x
⇒ f(100). f’(100) = 100
Illustration 20: If y = (^) ( )
lnx(^ lnx)(^ lnx)
(lnx)
∞ , then find dy dx
. (JEE MAIN)
Sol: Same as above
ln y = y ln(lnx)
1 y
× y’ = y xlnx
- ln(lnx).y ' ⇒ y’ (^1) ln(lnx) y
y xlnx
⇒ y’
1 yln(lnx) y
y xlnx
⇒ y’ =
y^2 (xlnx)(1 −ln(lnx)y)
Illustration 21: If log (x^2 + y^2 ) = 2 tan–
y x
, then prove that dy dx
x y x y
. (JEE ADVANCED)
Sol: Differentiating both the sides of the given relation with respect to x,
Mathematics | 21.
Illustration 24: If x = cosecθ – sinθ and y = cosecnθ – sin nθ, then find
dy dx.^ (JEE ADVANCED)
Sol: Differentiation of Parametric form.
x = cosec θ – sinθ
⇒ x^2 + 4 = (cosec θ – sin θ)^2 + 4 = (cosec θ + sin θ)^2 … (i)
and y^2 + 4 = (cosecnθ – sin nθ)^2 + 4 = (cosecnθ + sinnθ)^2 … (ii)
Now, dy dx
dy d dx d
(^) θ (^) θ
n(cosec^ n 1 )( cosec cot ) – nsinn 1 cos cosec cot cos
− (^) θ − θ θ − θ θ − θ θ − θ
n(cosec^ n cot sinn 1 cos (cosec cot cos )
θ θ + −θ θ θ θ + θ
ncot (cosec^ n sinn ) cot (cosec sin )
θ θ + θ θ θ + θ
n(cosec^ n sinn ) (cosec sin )
θ + θ θ + θ
Illustration 25: Find dy dx
if x = at^2 and y = 2at. (JEE MAIN)
Sol: Given that x = at^2 , y = 2at
Therefore, dx dt
= 2at and dy dt
= 2a.
Hence, dy dx
dy dt dx dt
= 2a 2at
t
Illustration 26: If x = cos 3 t and y = sin 3 t, then find dy dx
, for t 0, 2
(^) π ∈ ^
. (JEE MAIN)
Sol: dx dt = – 3 cos^2 t sint (≠ 0) ⇒ dy dt = 3 sin^2 t cos t ⇒ dy dx
dy dt dx dt
2 2
3 sin t cos t -3cos t sint
= – tan t
Illustration 27: If y = sec 4x and t = tan x, then prove that dy dt
4 2 4 2
16t(1 t ) (1 6t t )
. (JEE ADVANCED)
Sol: Write y in terms of t and differentiate.
y = 1 cos 4x
2 2
1 tan 2x 1 tan 2x
( )
( )
2 2
2 2
1 2t / (1 t )
1 2t / (1 t )
y =
2 2 2 2 2 2
(1 t ) 4t (1 t ) 4t
4 2 2 4 2 2
1 t – 2t 4t 1 t 2t 4t
4 2 4 2
1 t 2t 1 t 6t
dy dt
4 2 4 2
16t(1 t ) (1 6t t )
Illustration 28: If x = 1 2 lnt t
= 2x
2 dy dx
+ 1 (JEE MAIN).
Sol: Differentiation of Parametric form.
dy dx
= dy / dt dx / dt
21.14 | Methods of Differentiation and Applications of Derivatives
dy dt = (^ ) 2 t 0 2(1 / t) (3 2lnt) t
2 3 2lnt t
1 2lnt t
dx dt
(^2) ( ) 4
t 0 (1 / t) (1 lnt)2t t
= t^ 2t^4 2tlnt t
3
1 2 2lnt t
= – 1 2lnt 3 t
⇒ dy dx = t ⇒^ 2x
2 dy dx
10. DIFFERENTIATING WITH RESPECT TOANOTHER FUNCTION
Suppose u = f(x) and v = g(x) are two functions of x. To find the derivative of f(x) with respect to g(x), i.e. to find du dv
, the
formula du dv
= du / dx dv / dx
is used. Thus, to find the derivative of f(x) with respect tog(x), both are differentiated with
respect to x and then the derivative of f(x) with respect to x is divided by the derivative of g(x) with respect to x. The procedure is demonstrated in illustration 29.
Illustration 29: Differentiate sin 2 x with respect to ecosx^. (JEE MAIN)
Sol: Differentiate both the functions with respect to the common variable and use parametric form.
Let u(x) = sin^2 x and v(x) = ecosx. We want to find the value of du dv
= du / dx dv / dx
Clearly, du dx
= 2 sin x cos x and dv dx
= ecosx^ (– sin x) = – (sin x) ecosx
Hence, du dv
= 2sinx cos xcos x ( − sinx)e
= 2cos xcos x e
Illustration 30: Differentiate
2 2 2 2
1 x 1 x 1 x 1 x
with respect to 1 − x^4. (JEE ADVANCED)
Sol: Similar to the previous illustration.
Let u =
2 2 2 2
1 x 1 x 1 x 1 x
2 2 2 2 2
1 x 1 x (1 x ) – (1 x )
+^ +^ −
2 2 4 2
1 x 1 x 2 1 x 2x
⇒ u =
4 2
1 1 x x
du dx
2 3 4 4
4
x 0 ( 4x ) / (2 1 x ) 1 1 x 2x
x
^ ^ ^
⇒ du dx
5 4 4
4
( 2x ) / ( 1 x ) 2x 1 x 1
x
(^) ⇒ du dx
4 4 4
4 4
2x x^1 x^1 1 x 1 x x
− ^ ^ ^
− ^
4 4
2x (^) x 1 x 1 x x 1 x
4 4
2x (^1) x 1 x 1 x
….(i)
Let v = 1 −x^4
21.16 | Methods of Differentiation and Applications of Derivatives
derivatives are called as successive derivatives of y.
The following symbols are also used to denote the successive derivatives of y = f(x):
y 1 , y 2 , y 3 , ………. Yn, ………..
y’, y’’, y’’’, ………. yn, ……….. ⇒ dy dx
2 2
d y dx
3 3
d y dx
n n
d y dx
…….., ⇒ Dy, D^2 y, D^3 y, ……….Dny ………… (where D ≡ d dx
The following symbols are used to denote the value of the nth^ derivative at x = a.
yn(a), yn(a),
n n x a
d y dx (^) =
, Dny(a) & fn(a)
Misconception:
n^ n n
d y dy dx dx
Rohit Kumar (JEE 2012 AIR 79)
13. N th^ DERIVAVES OF SOME STANDARD FUNCTIONS
(a) Dn(ax + b)m^ = m(m – 1) (m –2) ……… (m – n + 1) an(ax + b)m-n
(b) If m ∈ N and m > n, thenDn(ax + b)m^ = m! (m − n)!
an(ax + b)m–n^ ; Dn(xm) = m! (m − n)!
xm–n
(c) Dn(ax + b)n^ = n! an; D n(xn) = n!
(d) Dn
ax b
n n n 1
( 1) n!a (ax b) +
; D n
x
n n 1
( 1) n! x +
(e) Dn{log(ax + b)} =
n 1 n
( 1) (n 1)! (ax b)
an; Dn(log x) =
n 1 n
( 1) (n 1)! x
(f) Dn(eax) = aneax
(g) Dn(amx) = mn^ (log a)n^ amx
(h) Dn{sin (ax + b)} = an^ sin ax b n 2
π +^ +
; Dn(sin x) = sin x n 2
π +
(i) Dn^ {cos (ax + b)} = ancos ax b n 2
(^) π +^ +
; D n^ (cos x) = cos x n 2
(^) π +
(j) D n^ {eax^ sin (bx + c)} = (a^2 + b^2 )n/2^ eax^ sin 1 bx c ntan b a
(k) Dn^ {eax^ cos (bx + c)} = (a^2 + b^2 )n/2^ eax^ cos 1 b bx c ntan a
(l) Dn^ tan 1 x a
n 1 n n
( 1) (n 1)!sin sinn a
− − − θ θ ,where θ = tan–
a x
(m) Dn^ (tan–1^ x) = (–1)n–1(n – 1)! sinnθ sin nθ, where θ = tan–
x
MASTERJEE CONCEPTS
Mathematics | 21.
14. LEIBNITZ THEOREM
If u and v are two functions such that their nth^ derivative exists, then the nth^ derivative of their product can be found by the following formula:
D n(uv) = (Dnu)v + nC 1 D n–1u. Dv + nC 2 D n–2u. D^2 v +……+……+nC (^) n-1Du. Dn-1v + ……..+ u.Dnv
The nth^ derivative of a product of two functions can be found out by using this theorem. While using this theorem, the second function in the product is the function whose successive derivative starts to vanish (if it is possible) after some stepsand the first function is a function whose nth^ derivative is easily known.
Illustration 32: If y = x 3 cos x, find Dny. (JEE MAIN)
Sol: Leibnitz theorem
Choose cos x as the first function and x^3 as the second function
Dn(cos x, x^3 ) = Dn(cos x) (x^3 ) + nC 1 Dn–1(cos x) (Dx^3 ) +nC 2 Dn–2(cos x). (D^2 x^3 ) + nC 3 Dn–3(cos x). (D^3 x^3 )
= x^3 cos x n 2
(^) π +
(^) − π +
− (^) 6x.cos x (n 2) 2
(^) − π +
− − (^) .6.cos x (n 3) 2
(^) + − π
= x^3 cos x n 2
(^) + π
(^) + π
(^) + π
- n(n – 1) (n – 2) sin x n 2
(^) + π
APPLICATION OF DERIVATIVES
1. THE INTERPRETATION OF THE DERIVATIVE
If y = f(x) be a given function, then the derivative/differential coefficient f’(x) or dy dx
at the point P(x 1 , y 1 ) is called the
trigonometric tangent of the angle ψ (say), which the positive direction of the tangent to the curve at P makes with
the positive direction of the x-axis. Therefore, dy dx
represents the slope of the tangent.
Thus, f’(x) = (x 1 ,y 1 )
dy dx
= Y Y
V
y=f(x) 90 o Tangent P(x , y ) 1 1 Normal
y
Figure 21.
Then,
(a) The inclination of the tangent with x-axis = tan–
dy dx
(b) Slope of the tangent = dy dx
(c) Slope of the normal = – dx dy
2. EQUATION OF TANGENT
(a) Equation of tangent to the curve y = f(x) at A(x 1 , y 1 ) is given by y – y 1 = (x 1 ,y 1 )
dy dx
(x – x 1 )
If the tangent at P (x 1 , y 1 ) of the curve y = f(x) is parallel to the x-axis (or perpendicular to the y-axis), then
Y = 0, i.e its slop will be equal to zero.
Mathematics | 21.
(c)
n n
x a
n n
y b
= 1,where x = a (sin θ)2/nand y = b(cos θ)2/n.
(d) For c 2 (x^2 + y^2 ) = x^2 y^2 , take x = c sec θ and y = c cosec θ.
(e) For y^2 = x^3 , take x = t^2 and y = t^3.
Illustration 33: If the tangent to the curve 2y^3 = ax 2 + x 3 at the point (a, a) cuts off intercepts α and β on the coordinate axes, where a^2 + b^2 = 61, the value of | a | is ____. (JEE MAIN)
(A) 16 (B) 28 (C) 30 (D) 31
Sol: (C) Write the equation of the tangent and find the value of α and β in terms of a. Then use a 2 + b^2 = 61 to find the value of a.
The slope of the tangent is given by dy dx
2 2
2ax 3x 6y
- (^) The value of this slope at (a, a) is 5/6.
Hence, the equation of tangent is y – a =
(x – a) ⇒ x −a / 5
y a / 6
Thus, the x-intercept α is – a 5
, and the y-intercept β is a 6
From a 2 + b^2 = 61, we get a^2 25
a^2 36
= 61 ⇒ a^2 = 25 × 36 ⇒ | a | = 30
3. EQUATION OF NORMAL
Equation of normal at (x 1 , y 1 ) to the curve y = f(x) isgiven by the following formula:
(y – y 1 ) =
(x 1 ,y 1 )
dy dx
(x – x 1 ) ⇒ (y – y 1 ) (x 1 ,y 1 )
dy dx
Some facts regarding the normal
(a) Slope of the normal drawn at point P (x 1 ,y 1 ) to the curve y = f(x) = – (x 1 ,y 1 )
dx dy
(b) If the normal makes an angle of θ with the positive direction of the x-axis, then – dx dy
= tanθ or dy dx
= – cotθ
(c) If the normal is parallel to the x-axis, then dx dy
= 0 or dy dx
(d) If the normal is parallel to the y-axis, then dx dy
= ∞ or dy dx
(e) If the normal is equally inclined from both the axes or cuts equal intercept, then – dx dy
= ± 1 or dy dx
(f) The length of the perpendicular from the origin to the normal is P’ =
1 1 2
dy x y dx
1 dy dx
21.20 | Methods of Differentiation and Applications of Derivatives
(g) The length of the intercept made by the normal on the x-axis is x 1 + y 1 dy dx
and the length of the intercept
on the y-axis is y 1 + x 1 dx dy
Illustration 34: Find out the distance between the origin and the normal to the curve y = e2x^ + x 2 at the point whose abscissa is 0. (JEE MAIN)
(A) 1 5
(B) 2
(C) 3
(D) 2
Sol: (B) Write the equation of the normal and find the distance of origin from the normal.
The point on the curve corresponding to x = 0 is (0, 1)
dy dx
= 2e2x^ + 2x ⇒ x 0
dy dx (^) =
Therefore, the equation of the normal at the point (0, 1) is
y – 1 = (– 1/2) (x – 0) ⇒ 2y + x – 2 = 0
Hence, the distance of the point (0, 0) from this line is 2 5
4. LENGTH OF TANGENT, NORMAL, SUBTANGENT AND SUBNORMAL
4.1 Tangent
PT = MP cosec Y = y 1 + cot^2 ψ =
2 y 1 dy dx dy dx
P
T O
y
G
y M
Figure 21.
4.2 Subtangent
TM = MP cot Y = y (dy / dx)
4.3 Normal
GP = MP sec Y = y 1 + tan^2 ψ =
2 y 1 dy dx
+ ^ ^
4.4 Subnormal
MG = MP tan (^) ψ = (^) y dy dx
Illustration 35: For the parabola y^2 = 16x, the ratio of the length of the subtangent to the abscissa is _____.
(A) 2 : 1 (B) 1 : 1 (C) X : Y (D) X^2 : Y (JEE MAIN)
