Derivative : The Chain Rule

Introduction

In many functions, a variable is inside another function — for example:

$$f(x)=(3x+2)^5$$

These are called composite functions, because one function is “inside” another.

To find their derivatives, we use a special rule called the Chain Rule.

The Chain Rule Formula

If: $$f(x)=u(v(x))$$

then the derivative is:

$$f^\prime(x)=\frac{d}{dx}f(x)=u^\prime(v(x))\cdot v^\prime(x)$$

example)

$$f(x)=(2x+1)^2=4x^2+4x+1$$ then

$$f^\prime(x)=4\cdot2x^{2-1}+4\cdot1x^{1-1}=8x+4$$ or

$$f^\prime(x)=2(2x+1)^{2-1}\cdot(2x+1)^\prime=4\cdot(2x+1)=8x+4$$

$$g(x)=(5x+9)^{10}$$ then

$$g^\prime(x)=10\cdot(5x+9)^{10-1}\cdot(5x+9)^\prime=10\cdot(5x+9)^9\cdot5=50(5x+9)^9$$

$$h(x)=(4x^2+5)^4$$ then

$$h^\prime(x)=4\cdot(4x^2+5)^{4-1}\cdot(4x^2+5)^\prime$$

$$=4\cdot(4x^2+5)^3\cdot(8x)=32x(4x^2+5)^3$$

$$p(x)=(6x^7+x^4)^5$$ then

$$p^\prime(x)=5\cdot(6x^7+x^4)^{5-1}\cdot(6x^7+x^4)^\prime$$

$$=5\cdot(6x^7+x^4)^4\cdot(42x^6+4x^3)=5(42x^6+4x^3)(6x^7+x^4)^4$$

$$q(x)=\sqrt{2x^2-5x+7}=(2x^2-5x+7)^\frac{1}{2}$$ then

$$q^\prime(x)=\frac{1}{2}\cdot(2x^2-5x+7)^{\frac{1}{2}-1}\cdot(2x^2-5x+7)^\prime$$

$$=\frac{1}{2}\cdot(2x^2-5x+7)^\frac{-1}{2}\cdot(4x-5)=\frac{4x-5}{2\sqrt{2x^2-5x+7}}$$

Chin rule

$$h(x)=f(g(x)) \longrightarrow h^\prime(x)=f^\prime(g(x))\cdot g^\prime(x)$$