# Derivative of a Function (Chain Rule) and Composite Function

## Derivative of a Function (Chain Rule) and Composite Function

Derivative of a Function (Chain Rule): If y is a differentiable function of t and t is a differentiable function of x i.e., y = f(t) and t = g(x)  then $$\frac{dy}{dx}\,=\,\frac{dy}{dt}.\frac{dt}{dx}$$.

Similarly, if y = f(u ), where u = g(v) and v =  h(x) then $$\frac{dy}{dx}\,=\,\frac{dy}{du}.\frac{du}{dv}.\frac{dv}{dx}$$.

Derivative of a Composite Function: If f(x) is differentiable at the point u = g(x) and g(x) is differentiable at x, then the composite function fog(x) = f{g(x)} is differentiable at x and  (fog)’(x) = f’{g(x)}.g’(x).

Example: Find the derivative of y = cosx³.sin²(x⁵) with respect to x.

Solution: Since the given function is a product of two different functions.

So using the product rule i.e., $$\frac{d}{dx}\,\left( uv \right)\,=\,u\,\frac{d}{dx}\,\left( v \right)\,+\,v\,\frac{d}{dx}\,\left( u \right)$$

On differentiating both sides w.r.t. x we get

$$\frac{dy}{dx}\,=\,\frac{d}{dx}$${cosx³ sin²(x⁵)}

= cosx³ $$\frac{d}{dx}$$ sin²(x⁵) + sin²(x⁵) $$\frac{d}{dx}$$ (cosx³)

$$\left[ U\sin g\,product\,rule,\,\frac{d}{dx}\,\left( uv \right)\,=\,u\,\frac{d}{dx}\,v\,+\,v\,\frac{d}{dx}\,u \right]$$

= (cos x³) (2 sin x⁵) $$\frac{d}{dx}$$ (sin x⁵) + sin² (x⁵) (-sin x³) $$\frac{d}{dx}$$ (x³)

$$\left[ U\sin g\,chain\,rule,\,\frac{d}{dx}\,f\left\{ g\left( x \right) \right\}\,=\,f’\left( x \right)\frac{d}{dx}\,g\left( x \right) \right]$$

= (cosx³) (2sinx⁵) (cosx⁵) $$\frac{d}{dx}$$ (x⁵) + sin²(x⁵) (-sinx³) (3x²)

= (cosx³) (2sinx⁵) (cosx⁵) (5x⁴) – sin²(x⁵) (-sinx³) (3x²)

= 10x⁴ (sinx⁵) (cosx⁵) (cosx³) – 3x²sinx³ sin²(x⁵)