# Differentiation – Some Standard Substitutions

## Differentiation – Some Standard Substitutions

Example: If $$\sqrt{1-{{x}^{6}}}+\sqrt{1-{{y}^{6}}}=a({{x}^{3}}-{{y}^{3}})$$, then prove that $$\frac{dy}{dx}=\frac{{{x}^{2}}}{{{y}^{2}}}\sqrt{\frac{1-{{y}^{6}}}{1-{{x}^{6}}}}$$.

Solution: Given that,

$$\sqrt{1-{{x}^{6}}}+\sqrt{1-{{y}^{6}}}=a({{x}^{3}}-{{y}^{3}})$$,

Let x³ = cosp and y³ = cosq

$$\sqrt{1-{{\cos }^{2}}p}+\sqrt{1-{{\cos }^{2}}q}=a(\cos p-\cos q)$$,

sinp + sin q = a(cosp – cosq)

$$2\sin \left( \frac{p+q}{2} \right)\cos \left( \frac{p-q}{2} \right)=-2a\sin \left( \frac{p-q}{2} \right)\sin \left( \frac{p+q}{2} \right)$$,

$$\tan \left( \frac{p-q}{2} \right)=-\frac{1}{a}$$,

$$p-q={{\tan }^{-1}}\left( -\frac{1}{a} \right)$$,

x³ = cosp and y³ = cosq

p = cos⁻¹ (x³) and q = cos⁻¹ (y³),

$${{\cos }^{-1}}({{x}^{3}})-{{\cos }^{-1}}({{y}^{3}})={{\tan }^{-1}}\left( -\frac{1}{a} \right)$$,

Differentiation with respect to x, we have

$$-\frac{3{{x}^{2}}}{\sqrt{1-{{x}^{6}}}}+\frac{3{{y}^{2}}}{\sqrt{1-{{y}^{3}}}}\frac{dy}{dx}=0$$,

$$\frac{dy}{dx}=\frac{{{x}^{2}}}{{{y}^{2}}}\sqrt{\frac{1-{{y}^{3}}}{1-{{x}^{3}}}}$$.