**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.