# Differentiation – Rules 2

## Differentiation – Rules 2

Derivative of Composite Function: If f(u) 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).

Relation between dy/dx and dx/dy: If the inverse functions f and g are defined by y = f(x) and x = g(y) and if f’(x) exists and f’(x) ≠ 0, then $$g'(y)=\frac{1}{f'(x)}$$. This result can also be written as, if $$\frac{dy}{dx}$$ exists and $$\frac{dy}{dx}\ne 0$$, then

$$\frac{\frac{dx}{dy}=1}{\left( \frac{dy}{dx} \right)}or\frac{dy}{dx}.\frac{dx}{dy}=1$$,

Or $$\frac{\frac{dy}{dx}=1}{\left( \frac{dx}{dy} \right)},\,where\left[ \frac{dx}{dy}\ne 0 \right]$$,

Note: Differentiation of an even function is an odd function and differentiation of an odd function is an even function.

Example: Derivative of cos x³. sin² (x⁵) with respect to x is

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

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

Let y = cosx³ sin²(x⁵)

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) $$\frac{d}{dx}$$ (cos x³)

[using product rule, $$\frac{d}{dx}$$ (uv) = u$$\frac{d}{dx}$$v + v$$\frac{d}{dx}$$ u]

= (cos x³)(2sin x⁵) $$\frac{d}{dx}$$ (sin x⁵) + sin² (x⁵)(-sinx³) $$\frac{d}{dx}$$ (x³)

[using chain rule, $$\frac{d}{dx}$$f{g(x)} = f’(x) $$\frac{d}{dx}$$g(x)]

= (cos x³)(2 sin x⁵)(cos x⁵) $$\frac{d}{dx}$$ (x⁵) + sin² (x⁵) (-sinx³)(3x²) (using chain rule)

= (cos x³) (2 sin x⁵)(cos x⁵)(5x⁴) – sin²(x⁵)(sin x³)(3x²)

= 10x⁴ (sin x⁵)(cos x⁵)(cos x³) – 3x² sinx³ sin²x⁵.