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DERIVATIVES OF TRIGONOMETRIC FUNCTIONS.
- Show that the derivatives of 𝑓(𝑥) = sin 𝑥 is 𝑐𝑜𝑠 𝑥.
Solution
We first note that;
(𝑠𝑖𝑛 (𝑥 + ℎ)) = 𝑠𝑖𝑛 𝑥 𝑐𝑜𝑠ℎ + 𝑐𝑜𝑠𝑥 𝑠𝑖𝑛 ℎ
Let 𝑓(𝑥) = 𝑠𝑖𝑛 𝑥
Then,
𝑓′ (𝑥) = lim ℎ→0𝑓(𝑥+ℎ)−𝑓(𝑥)ℎ
== lim ℎ→0 sin(𝑥+ℎ)−sin 𝑥ℎ
= (^) ℎ→0lim (sin 𝑥 cos ℎ + 𝑐𝑜𝑠𝑥 sin ℎ − 𝑠𝑖𝑛𝑥ℎ )
=lim ℎ→0 (sin 𝑥(cos ℎ−1)+cos 𝑥 𝑠𝑖𝑛 ℎℎ )
= (^) ℎ→0lim (sin 𝑥 .cos ℎ−1ℎ ) + lim ℎ→0 (cos 𝑥. 𝑠𝑖𝑛ℎℎ )
= sin 𝑥. lim ℎ→0 (cos ℎ−1ℎ ) + cos 𝑥. lim ℎ→0 (sin ℎℎ ) { (^) ℎ→0lim (sin ℎℎ ) = 1; lim ℎ→0 (cos ℎ−1ℎ ) = 0}
=sin 𝑥. (0) + cos 𝑥. (1)
= cos 𝑥
Hence 𝑑𝑦𝑑𝑥 (sin 𝑥) = cos 𝑥
- Show that (^) 𝑑𝑥𝑑 (cos 𝑥) = − sin 𝑥
Note that; cos(𝑥 + ℎ) = cos 𝑥 𝑐𝑜𝑠ℎ − sin 𝑥 sin ℎ
Solution
𝑑 𝑑𝑥 (cos 𝑥) = lim ℎ→ cos(𝑥 + ℎ) ℎ − cos 𝑥
= (^) ℎ→0limcos 𝑥 cos ℎ−sin 𝑥 𝑠𝑖𝑛ℎℎ − cos 𝑥
= ℎ→0limcos 𝑥 (cos ℎ−1)−𝑠𝑖𝑛𝑥 sin ℎℎ
= ℎ→0limcos 𝑥 .𝑐𝑜𝑠ℎ−1ℎ − lim ℎ→0sin 𝑥 𝑠𝑖𝑛ℎℎ
= cos 𝑥. lim ℎ→0𝑐𝑜𝑠ℎ−1ℎ − 𝑠𝑖𝑛𝑥. lim ℎ→0sin ℎℎ {lim ℎ→0sin ℎℎ = 1; (^) ℎ→0limcosh −1ℎ = 0}
= cos 𝑥. (0) − sin 𝑥. (1)
Therefore, (^) 𝑑𝑥𝑑 (cos 𝑥) = − sin 𝑥
Examples
Find the derivatives of the following functions.
- 𝑦 = 𝑥^2 − sin 𝑥
Solution
𝑑𝑦 𝑑𝑥 =^
𝑑 𝑑𝑥 (𝑥
(^2) − sin 𝑥) = 2𝑥 − 𝑑 𝑑𝑥(sin x) = 2𝑥 − 𝑐𝑜𝑠 𝑥
- 𝑦 = 𝑥^2 𝑠𝑖𝑛𝑥
Solution
𝑦 = 𝑥^2 𝑠𝑖𝑛𝑥
𝑑𝑦 𝑑𝑥
(𝑥^2 𝑠𝑖𝑛𝑥)
= 𝑥^2 𝑑𝑥𝑑 (𝑠𝑖𝑛𝑥) + 2𝑥 𝑠𝑖𝑛𝑥
= 𝑥^2 cos 𝑥 + 2𝑥 𝑠𝑖𝑛𝑥
Let 𝑦 = (^) 1−𝑠𝑖𝑛𝑥cos 𝑥 = 𝑢𝑣
𝑑𝑦 𝑑𝑥 =
𝑣𝑢′^ − 𝑢𝑣′
𝑣^2
(1−sin 𝑥) (^) 𝑑𝑥𝑑(𝑐𝑜𝑠𝑥)−cos 𝑥 (^) 𝑑𝑥𝑑(1−𝑠𝑖𝑛𝑥) (1−sin 𝑥)^2
= (1−sin 𝑥)(− sin 𝑥)−𝑐𝑜𝑥 𝑥(0−cos 𝑥)(1−sin 𝑥) 2
= − sin 𝑥+ 𝑠𝑖𝑛
(^2) 𝑥+ 𝑐𝑜𝑠 (^2) 𝑥 (1−sin 𝑥)^2
= (^) (1−sin 𝑥)(1−𝑠𝑖𝑛𝑥) 2 = (^) (1−𝑠𝑖𝑛𝑥)^1
- 𝑦 = sin 5𝑥
Solution
𝑦 = sin 5𝑥 = sin 𝑢 𝑢 = 5𝑥
𝑑𝑦 𝑑𝑥
(sin 5𝑥) =
sin 𝑢
𝑑𝑦 𝑑𝑥
= cos 𝑢 5𝑥 = (cos 5𝑥) = 5(𝑐𝑜𝑠 5𝑥)
- 𝑦 = (^) 1+cos 𝑥sin 𝑥
Solution
Let 𝑦 = (^) 1+cos 𝑥sin 𝑥 = 𝑢𝑣
𝑢 = 𝑠𝑖𝑛 𝑥 ; 𝑑𝑢 = cos 𝑥 = 𝑢′
𝑣 = 1 + 𝑐𝑜𝑠 𝑥 ; 𝑑𝑣 = − sin 𝑥 = 𝑣′
Therefore,
𝑑𝑦 𝑑𝑥
𝑣^2
(1 + cos 𝑥) cos 𝑥 − sin 𝑥(− sin 𝑥) (1 + cos 𝑥)^2
= cos 𝑥+ 𝑐𝑜𝑠
(^2) 𝑥+𝑠𝑖𝑛 (^2) 𝑥 (1+cos 𝑥)^2 =^
cos 𝑥+ (1+cos 𝑥)^2 =^
1 1+cos 𝑥
DERIVATIVES OF OTHER TRIGONOMETRIC FUNCTIONS
As 𝑠𝑖𝑛 𝑥 and 𝑐𝑜𝑠 𝑥 are differentiable function of 𝑥, then related functions are differentiable at every value of 𝑥
𝑡𝑎𝑛 𝑥 = (^) cos 𝑥sin 𝑥 ; cot 𝑥 = cos 𝑥sin 𝑥 ; sec 𝑥 = (^) cot 𝑥^1 ; csc 𝑥 = (^) sin 𝑥^1
These derivatives calculated from quotient rule are given by the following formulas,
- (^) 𝑑𝑥𝑑 = (tan 𝑥) = 𝑠𝑒𝑐^2 𝑥 2. (^) 𝑑𝑥𝑑 (cot 𝑥) = −𝑐𝑠𝑐^2 𝑥
- (^) 𝑑𝑥𝑑 = (sec 𝑥) = sec 𝑥 tan 𝑥 4. (^) 𝑑𝑥𝑑 (csc 𝑥) = − csc 𝑥 cot 𝑥
Exercise: show 1-4 above
- (^) 𝑑𝑥𝑑 = (tan 𝑥) = 𝑠𝑒𝑐^2 𝑥
Solution
Let 𝑦 = tan 𝑥 = sin 𝑥cos 𝑥 = 𝑢𝑣
𝑢′^ = cos 𝑥 , 𝑣′^ = − sin 𝑥
𝑑𝑦 𝑑𝑥
(tan 𝑥) =
sin 𝑥 𝑐𝑜𝑠 𝑥
′−𝑢𝑣′ 𝑣^2 =^
(cos 𝑥) cos 𝑥−(sin 𝑥)(−𝑠𝑖𝑛𝑥) 𝑐𝑜𝑠^2 𝑥
= 𝑐𝑜𝑠
(^2) 𝑥+ 𝑠𝑖𝑛 (^2) 𝑥 𝑐𝑜𝑠^2 𝑥 =^
1 𝑐𝑜𝑠^2 𝑥 = (^
1 cos 𝑥)
2 = 𝑠𝑒𝑐^2 𝑥
- (^) 𝑑𝑥𝑑 (csc 𝑥) = − csc 𝑥 cot 𝑥
Solution
Let csc 𝑥 = (^) sin 𝑥^1 = 𝑢𝑣
𝑢 = 1, 𝑢′^ = 0, 𝑣 = sin 𝑥, 𝑣′^ = cos 𝑥
Example 1
Assuming y to be a function of x. find 𝑑𝑦𝑑𝑥 for 𝑥^2 + 𝑦^2 = 4
Solution
𝑑 𝑑𝑥
(𝑥^2 + 𝑦^2 ) =
(𝑥^2 ) +
(𝑦^2 ) =
This process is called implicit differentiation.
Example 2
Find 𝑑𝑦𝑑𝑥 for 𝑥^3 + 3𝑥^2 𝑦 + 𝑦^3 = 1
Solution
𝑑 𝑑𝑥
(𝑥^3 + 3𝑥^2 𝑦 + 𝑦^3 ) =
(𝑥^3 ) + 3
(𝑥^2 𝑦) +
𝑦^3 ) =
3𝑥^2 + 3 (𝑦. 6𝑥 + 3𝑥^2
) + 3𝑦^2
3𝑥^2 + 6𝑥𝑦 + 3𝑥^2
+ 3𝑦^2
(𝑥^2 + 𝑦^2 ) = −3𝑥^2 − 6𝑥𝑦
𝟑(𝒙𝟐^ + 𝒚𝟐)
𝒙𝟐^ + 𝒚𝟐
Example 3
Find the coordinates of the point on the graph of (𝑥 − 2𝑦 − 1)^2 + (𝑥 + 𝑦)^2 = 16……………..(1)
Where the tangent line is horizontal.
Solution
The tangent line is horizontal at the point where 𝑑𝑦𝑑𝑥 = 0
Solving for 𝑑𝑦𝑑𝑥
𝑑𝑦 𝑑𝑥 = (−2𝑥 + 10𝑦 + 4) + (4𝑥 − 2𝑦 − 2) = 0
𝑑𝑦 𝑑𝑥 =
Substituting 𝑦 = 2𝑥 − 1 into the first equation we get
(𝑥 − 2(2𝑥 − 1) − 1)^2 + (𝑥 + (2𝑥 − 1))^2 = 16
To get
the points where the tangent is horizontal (1.27, 1.55) and (-0.609,-2.22)
DERIVATIVES OF PARAMETRIC EQUATIONS
Up to now all the equations we have dealt with have been in the form 𝑦 = 𝑓(𝑥). In either case, a direct relationship between x and y is given.
However sometimes it is more convenient to express both x and y in terms of a 3rd^ variable called a parameter. e.g. 𝑥 = 𝑓(𝑡), 𝑦 = 𝑓(𝑡).
For instant consider the parametric equation 𝑥 = 𝑠𝑖𝑛 𝑡 𝑎𝑛𝑑 𝑦 = 𝑐𝑜𝑠 𝑡, 𝑖𝑓 𝑡 = 0, 𝑡ℎ𝑒𝑛 𝑥 = 0 𝑎𝑛𝑑 𝑦 = 1
Example
Circle
Given s circle with centre at the origin and radius r. denote by 𝜃 the angle formed by x-axis and radius to some point 𝑚(𝑥, 𝑦) of the circle. Then the coordinates of any point on the circle with be expressed in terms of the parameter 𝜃 as follows.
𝑥 = 𝑟𝑐𝑜𝑠 𝜃
𝑦 = 𝑟𝑠𝑖𝑛 𝜃; 0 ≤ 𝜃 ≤ 2𝜋
These are parametric equations of the circle, we can eliminate the parameter from these equations to get the equation of a circle in x and y.
i.e. 𝑥^2 + 𝑦^2 = (𝑟𝑐𝑜𝑠 𝜃)^2 + (𝑟𝑠𝑖𝑛 𝜃)^2
= 𝑟^2 𝑐𝑜𝑠^2 𝜃 + 𝑟^2 sin^2 𝜃
=𝑟^2 (𝑐𝑜𝑠^2 𝜃 + sin^2 𝜃)
=𝑟^2 (1) = 𝑟^2
𝑥^2 + 𝑦^2 = 𝑟^2
Suppose we are interested in finding the slope of a graph that is represented in parametric form. i.e. finding 𝑑𝑦𝑑𝑥 we shall use the following theorem.
Theorem 1
𝑖𝑓 = 𝑓(𝑡), 𝑦 = 𝑔(𝑡) 𝑎𝑛𝑑
Example 1
Find 𝑑𝑦𝑑𝑥 if 𝑥 = 𝑡^2 + 5t − 9, y = t + 3
Solution
𝑥 = 𝑡^2 + 5𝑡 + 9 ; 𝑑𝑥𝑑𝑡 = 2𝑡 + 5;
Suppose we want to take a second derivative. The first derivative is a function of the parameter t, while its derivatives taken w.r.t x.
By chain rule
𝑑^2 𝑦 𝑑𝑥^2 =
Where as 𝑑
(^3) 𝑦 𝑑𝑥^3 =
𝑑 𝑑𝑥(
𝑑^2 𝑦 𝑑𝑥^2 ) 𝑑𝑥 𝑑𝑡
Since, (^) 𝑑𝑡𝑑 (𝑑𝑦𝑑𝑥) = (^) 𝑑𝑥𝑑 (𝑑𝑦𝑑𝑥). 𝑑𝑥𝑑𝑡 = 𝑑
(^2) 𝑦 𝑑𝑥^2.
𝑑𝑥 𝑑𝑡
From the above example.
𝑑^2 𝑦 𝑑𝑥^2 =
𝑑𝑡 (^
− (2𝑡+5). 2 =^
− (2𝑡+5)^3
Example 2
Find 𝑑𝑦𝑑𝑥 𝑎𝑛𝑑 𝑑
(^2) 𝑦 𝑑𝑥^2 𝑓𝑜𝑟 𝑥 = 𝑡
Solution
DERIVATIVES OF HIGHER ORDER
The derivatives of a function is also a function e.g. 𝑓(𝑥) = 𝑥^4 then 𝑓′(𝑥) = 4𝑥^3.
Since 𝑓′(𝑥)^ is itself a function we can find its derivatives (if it exists) the results called the second derivative of 𝑓 written as 𝑓′′(𝑥). Thus 𝑓′′(𝑥) = 12𝑥^2 for the above example.
We may continue indefinitely, taking the 3 𝑟𝑑^ , 4𝑡ℎ𝑎𝑛𝑑 5𝑡ℎ^ 𝑤ℎ𝑖𝑐ℎ 𝑎𝑟𝑒 𝑤𝑟𝑖𝑖𝑡𝑒𝑛 𝑎𝑠 𝑓𝐼𝐼, 𝐹𝐼𝑉𝑎𝑛𝑑 𝑓𝑉
For the above example 𝑓𝐼𝐼𝐼(𝑥) = 24𝑥 , 𝑓𝐼𝑉(𝑥) = 24 𝑎𝑛𝑑 𝑓𝑣(𝑥) = 0
Example 1
Find the first three derivatives of 𝑦 = 𝑥^2 + 𝑥1/
Solution
𝑦𝐼^ =
−1/
𝑦𝐼𝐼^ =
𝑑^2 𝑦
𝑑𝑥^2
3 (^2) = 2 − 1/4𝑥−3/
𝑦𝐼𝐼𝐼^ =
𝑑^3 𝑥
𝑑𝑥^3
5 (^2) =^3 8
5 2
Exercise
Find 𝑑
(^2) 𝑦 𝑑𝑥^2 𝑓𝑜𝑟 𝑥
DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS
In this section we are going to look at the derivatives of inverse trigonometric functions which are derived as 𝑎𝑟𝑐 𝑠𝑖𝑛 𝑥, 𝑎𝑟𝑐 𝑡𝑎𝑛 𝑥, 𝑎𝑟𝑐 𝑐𝑜𝑡 𝑥 𝑎𝑟𝑐 𝑠𝑒𝑐 𝑥 𝑎𝑛𝑑 𝑎𝑟𝑐 𝑐𝑠𝑒𝑐 𝑥
Inverse function exists when appropriate restrictions are placed on the domain of the original functions. E.g. the domain for arc sin x is from -1 to 1, the output for arc sin x is all angles from –π/ to π/2 radians.
The domain of trigonometric functions restricted appropriately so that they become one to one function and the inverse can be determined.
Examples
- Find the derivative of arc sin x i.e. 𝑦 = 𝑠𝑖𝑛−1𝑥
Solution
𝑦 = 𝑠𝑖𝑛−1𝑥==> 𝑥 = sin 𝑦 𝑑 𝑑𝑥
(sin 𝑦)
1 = cos 𝑦.
cos 𝑦
𝑏𝑢𝑡 cos 𝑦 = √1 − 𝑠𝑖𝑛^2 𝑥 = √1 − 𝑥^2 𝑑𝑦 𝑑𝑥 =^
cos 𝑦 =^
√1 − 𝑥^2
- Find the 𝑎𝑟𝑐 𝑐𝑜𝑠 𝑥 𝑜𝑟 𝑦 = 𝑐𝑜𝑠−1^ 𝑥 Solution Let 𝑦 = 𝑐𝑜𝑠−1𝑥 == cos 𝑦 = 𝑥 Therefore, (^) 𝑑𝑥𝑑 (cos 𝑦) = (^) 𝑑𝑥𝑑 (𝑥) = − sin 𝑦 𝑑𝑦𝑑𝑥 = 1
Below are the derivatives of six inverse trigonometric function that can be derived by a similar process as we did above.
- (^) 𝑑𝑥𝑑(𝑠𝑖𝑛−^1 𝑥)= (^) √ 11 −𝑥 2 −^1 ≤^ 𝑥^ ≤^1
- (^) 𝑑𝑥𝑑(𝑐𝑜𝑠−^1 𝑥)= (^) √ 1 −−^1 𝑥 2 −^1 ≤^ 𝑥^ ≤^1
3. 𝑑𝑥𝑑(𝑡𝑎𝑛−^1 𝑥)= 1 +^1 𝑥 2 −∞^ ≤^ 𝑥^ ≤^ ∞
4. 𝑑𝑥𝑑(𝑐𝑜𝑡−^1 𝑥)= 1 −+^1 𝑥 2 −∞^ ≤^ 𝑥^ ≤^ ∞
5. 𝑑𝑥𝑑(𝑐𝑠𝑐−^1 𝑥)= |𝑥|√−𝑥^12 − 1
6. 𝑑𝑥𝑑(𝑠𝑒𝑐−^1 𝑥)= |𝑥|√^1 𝑥 2 − 1
Examples
Given that 𝑦 = 𝑠𝑖𝑛−1(2𝑥) find 𝑑𝑦𝑑𝑥
Solution
Let 2𝑥 = 𝑡 = 𝑦 = 𝑠𝑖𝑛−1(𝑡)
𝑡 = sin 𝑦 𝑑𝑡 𝑑𝑥 = 2,^
𝑑 𝑑𝑡 (𝑡) =^
𝑑 𝑑𝑡 (sin 𝑦) 1 = 𝑐𝑜𝑠𝑦.
𝑑𝑡 =^
cos 𝑦
Therefore 𝑑𝑦𝑑𝑥 = 𝑑𝑦𝑑𝑡. (^) 𝑑𝑥𝑑𝑡 = (^) cos 𝑦^1. 2 = ( (^) √1−𝑠𝑖𝑛^12 𝑦). 2 = (^) √1−𝑥^22
√1 − 𝑢^2
√1 − (2𝑥)^2
𝑑𝑥 (2𝑥) =^
√1 − 4𝑥^2
- Find 𝑑𝑦𝑑𝑥 𝑜𝑓 𝑦 = 𝑡𝑎𝑛−1(1 + 𝑥^2 )
If 𝑢 = 1 + 𝑥^2 == 𝑑𝑢𝑑𝑥 = 2𝑥
Therefore (^) 𝑑𝑥𝑑 [𝑡𝑎𝑛−1(1 + 𝑥^2 )] = (^) 1+(1+𝑥^12 ) 2. (^) 𝑑𝑥𝑑 (1 + 𝑥^2 ) = (^) 𝑥 (^4) +2𝑥2𝑥 (^2) +
- 𝑦 = 𝑥^3 ln(2𝑥 + 50) 𝑙𝑒𝑡 𝑢 = 𝑥^3 , 𝑣 = ln(2𝑥 + 50) 𝑢′^ = 3𝑥^2 , 𝑣′^ =
(𝑢, 𝑣) = 𝑢𝑣′^ + 𝑢′𝑣
𝑥^3 (2)
2𝑥^3
LOGARITHMIC DIFFERENTIATION
In determining the derivatives of certain kind of functions it is considerably advantage to take the logarithm of a function on both sides before differentiating. Example
- A function of the form 𝑦 = 𝑓(𝑥)𝑔(𝑥) where both f and g are non-constants.
- Derivatives of quotients can also be determines by this method. e.g. 𝑦 = 𝑢𝑣 , 𝑤ℎ𝑒𝑟𝑒 𝑢 𝑎𝑛𝑑 𝑣 𝑎𝑟𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑥 i.e. ln 𝑦 = ln (
𝑣) = ln 𝑢 − ln 𝑣 𝑑 𝑑𝑢 (ln 𝑦) =
𝑑𝑢 ln 𝑢 −
𝑑𝑥 ln 𝑣 1 𝑦
𝑑𝑦 𝑑𝑥 = 𝑦(
1 𝑢
𝑑𝑢 𝑑𝑥 −^
1 𝑣
𝑑𝑣 𝑑𝑥)=^
𝑢 𝑣 (
1 𝑢
𝑑𝑢 𝑑𝑥 −^
1 𝑣
𝑑𝑣 𝑑𝑥) =^
𝑣𝑑𝑢𝑑𝑥−𝑢𝑑𝑥𝑑𝑣 𝑣^2 which is quotient rule.
Examples
- Find the derivatives of the following 𝑦 = 𝑓(𝑥) = 𝑥𝑥 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 ln 𝑦 = ln 𝑥𝑥^ = 𝑥 ln 𝑥 ln 𝑦 =
𝑑𝑥 = ln 𝑥 + 𝑥.
Therefore 𝑑𝑦𝑑𝑥 = 𝑦(ln 𝑥 + 1) = 𝑥𝑥(ln 𝑥 + 1)
- 𝑦 = 2sin 𝑥 𝑑𝑦 𝑑𝑥 =
sin 𝑥) = (2sin 𝑥 (^) ln 2) 𝑑 𝑥 (sin 𝑥) = 2sin 𝑥^ ln 2 cos 𝑥
- 𝑦 = 𝑥sin 𝑥 ln 𝑦 = ln 𝑥sin 𝑥^ = sin 𝑥 ln 𝑥 𝑑 𝑑𝑥 (ln 𝑦) =
𝑑𝑥 (sin 𝑥 ln 𝑥) 1 𝑦
− cos 𝑥 ln 𝑥 +
sin 𝑥 𝑑𝑦 𝑑𝑥 = 𝑦(cos 𝑥𝑙𝑛 𝑥 +
sin 𝑥) 𝑑𝑦 𝑑𝑥 = 𝑥sin 𝑥(cos 𝑥𝑙𝑛 𝑥 +
sin 𝑥)
DERIVATIVES OF EXPONETIAL FUNCTIONS
Find the derivatives of 𝑦 = 𝑎𝑥, 𝑎 > 0 Solution 𝑦 = 𝑎𝑥 ln 𝑦 = ln 𝑎𝑥 𝑑 𝑑𝑥 (ln 𝑦) =
= ln 𝑎 𝑑𝑦 𝑑𝑥 = 𝑦 ln 𝑎 Therefore, (^) 𝑑𝑥𝑑 (𝑎𝑥) = 𝑎𝑥^ ln 𝑎 If u is a differentiable function of x, then 𝑑 𝑑𝑥 (𝑎
𝑢) = 𝑎𝑢 (^) ln 𝑎. 𝑑𝑢 𝑑𝑥 If 𝑎 = 𝑒 Euler number so that, 𝑦 = 𝑒𝑥, we obtain;
𝑑 𝑑𝑥 (𝑒𝑥^ ) = 𝑒𝑥^. ln 𝑒 = 𝑒𝑥^ =
If u is a differentiable function of x then, 𝑦 = 𝑒𝑢 𝑑𝑦 𝑑𝑥
(𝑒𝑢) = 𝑒𝑢^ ln 𝑒.
= 𝑒𝑢^
