Subject: Calculus

# Chain Rule

Single functions are easily differentiated. But a combination of two or more functions which is called a composite function sometimes possesses difficulty. Well, not all the times. Thanks to the "Chain Rule". The following theorem is commonly known as chain rule. They apply to functions that are composite.

## Theorem: (Derivative of a Composite Function)

Supposet f(x)=[h(x)]^{r}, where r is any rational number and h(x) is a function of x. Then the derivative of f(x) is given by

 f'(x)=r[h(x)]^{r-1}\cdot [h'(x)].

### Example #1

Suppose that f(x)=[2x]^{5}-3. Find the derivative of f.

Solution: By the above theorem, it is clear that

 f'(x) = 5[2x]^{5-1}\cdot D_{x}(2x)-D_{x}(3) = 5(2x)^{4}\cdot 2-0 = 10(2x)^{4}

Thus, the derivative of f(x)=[2x]^{5}-3 is 20x^{4}.

10(2x)^{4}

### Example #2

Suppose that g(x)=(6x-3)^{-3}+7x. Find g'(x).

Solution: By the above theorem, it is clear that

 g'(x) = -3(6x-3)^{-3-1}\cdot D_{x}(6x-3)+D_{x}(7x) = -3(6x-3)^{-4}\cdot 6-7 = 18(6x-3)^{-4}-7

Thus, the derivative of g(x)=(6x-3)^{-3}+7x is -18(6x-3)^{-4}-7.

-18(6x-3)^{-4}-7

### Example #3

Suppose that s(x)=(ax)^{b}+\frac{c}{x^{d}}. Find \dot{s}.

Solution: Note that the second term of the above example may be applied with the quotient rule as discussed earlier. We can do that but also, we can make \frac{c}{x^{d}}=cx^{-d}. Thus, according to the above theorem, the derivative is given by

 \dot{s}=s'(x) = b(ax)^{b-1}\cdot D_{x}(ax)+c\cdot (-d)\cdot x^{-d-1}D_{x}(x) = ab(ax)^{b-1}-cdx^{-(d+1)} = ab\frac{(ax)^{b}}{ax}-\frac{cd}{x^{d+1}}.

Accordingly, the derivative of s(x)=(ax)^{b}+\frac{c}{x^{d}} is given by ab\frac{(ax)^{b}}{ax}-\frac{cd}{x^{d+1}}.

ab\frac{(ax)^{b}}{ax}-\frac{cd}{x^{d+1}}

### Example #4

Suppose that f(x)=e^{\cos{x}}. Find \frac{d}{dx}f.

Solution:

 \frac{d}{dx}f=f'(x) = e^{\cos{x}}\cdot D_{x}\cos{x} = -\sin{x}e^{\cos{x}}

Thus,

-\sin{x}e^{\cos{x}}

### Example #5

Suppose that g(x)=\sin{e^{2x^2}}. Find the first derivative of g.

Solution:

 \dot{g}=g'(x) = \cos{e^{2x^2}}\cdot D_x e^{2x^2} = \cos{e^{2x^2}}\cdot 2e^{2x^2}\cdot D_x x^2 = \cos{e^{2x^2}}\cdot 2e^{2x^2}\cdot 2x = \cos{e^{2x^2}}\cdot 4xe^{2x^2} .

Thus,