Differentiation by Chain Rule

1. Concept Overview

If a function is a composition of two (or more) functions, say [y = f(g(x))], then the derivative is found by differentiating the outer function at the inner function and multiplying by the derivative of the inner function.

2. Formal Rule (Chain Rule)

If [y = f(g(x))] and both [f] and [g] are differentiable, then

[ \dfrac{dy}{dx} = f'(g(x))\cdot g'(x) ].

In words: “Derivative of outer (evaluated at inner) × derivative of inner.”

You can extend this to three-level composition: if [y = f(g(h(x)))], then

[ \dfrac{dy}{dx} = f'(g(h(x)))\cdot g'(h(x))\cdot h'(x) ].

3. Worked Examples (Step-by-step)

Example 1 — Polynomial inside polynomial

Differentiate [f(x) = (3x^{2} + 2x + 1)^{5}].

Step-by-Step Solution

1. Inner function: [g(x)=3x^{2}+2x+1]
   Outer function: [F(u)=u^{5}]
2. [\dfrac{d}{du}(u^{5}) = 5u^{4}]
3. [\dfrac{d}{dx}(3x^{2}+2x+1)][ = 6x+2]
4. Apply chain rule:
   [\dfrac{d}{dx}\big((3x^{2}+2x+1)^{5}\big)][ = 5(3x^{2}+2x+1)^{4}(6x+2)]

Conclusion:
[\dfrac{d}{dx}\big((3x^{2}+2x+1)^{5}\big)][ = 5(3x^{2}+2x+1)^{4}(6x+2)]

Example 2 — Trigonometric outer, linear inner

Differentiate [f(x)=\sin(4x+1)].

Step-by-Step Solution

1. Inner: [g(x)=4x+1]
   Outer: [F(u)=\sin(u)]
2. [\dfrac{d}{du}(\sin u)=\cos u]
3. [\dfrac{d}{dx}(4x+1)=4]
4. Apply chain rule:
   [\dfrac{d}{dx}\big(\sin(4x+1)\big)][ = \cos(4x+1) \cdot 4]

Conclusion:
[\dfrac{d}{dx}\big(\sin(4x+1)\big)][ = 4\cos(4x+1)]

Example 3 — Exponential of a quadratic

Differentiate [f(x)=e^{x^{2}}].

Step-by-Step Solution

1. Inner: [g(x)=x^{2}]
   Outer: [F(u)=e^{u}]
2. [\dfrac{d}{du}(e^{u})=e^{u}]
3. [\dfrac{d}{dx}(x^{2})=2x]
4. Chain rule:
   [\dfrac{d}{dx}\big(e^{x^{2}}\big)][ = e^{x^{2}}\cdot 2x]

Conclusion:
[\dfrac{d}{dx}\big(e^{x^{2}}\big)][ = 2xe^{x^{2}}]

Example 4 — Square root (power) of a function

Differentiate [f(x)=\sqrt{1+3x}][ = (1+3x)^{1/2}], domain [(1+3x)>0].

Step-by-Step Solution

1. Inner: [g(x)=1+3x]
   Outer: [F(u)=u^{1/2}]
2. [\dfrac{d}{du}(u^{1/2})][ = \dfrac{1}{2u^{1/2}}]
3. [\dfrac{d}{dx}(1+3x)=3]
4. Chain rule:
   [\dfrac{d}{dx}\big((1+3x)^{1/2}\big)][=\dfrac{1}{2(1+3x)^{1/2}}\cdot 3]
5. Simplify:
   [\dfrac{3}{2\sqrt{1+3x}}]

Conclusion:
[\dfrac{d}{dx}\big(\sqrt{1+3x}\big)][ = \dfrac{3}{2\sqrt{1+3x}}]

Example 5 — Composition with inverse trig inside a polynomial (two-level)

Differentiate [f(x)=\tan^{-1}(x^{3})], domain all real.

Step-by-Step Solution

1. Inner: [g(x)=x^{3}]
   Outer: [F(u)=\tan^{-1}(u)]
2. [\dfrac{d}{du}(\tan^{-1}u)][ = \dfrac{1}{1+u^{2}}]
3. [\dfrac{d}{dx}(x^{3}) = 3x^{2}]
4. Chain rule:
   [\dfrac{d}{dx}\big(\tan^{-1}(x^{3})\big)][ = \dfrac{1}{1+x^{6}} \cdot 3x^{2}]
5. Final result:
   [\dfrac{3x^{2}}{1+x^{6}}]

Conclusion:
[\dfrac{d}{dx}\big(\tan^{-1}(x^{3})\big)][ = \dfrac{3x^{2}}{1+x^{6}}]