Differentiation – Some Standard Substitutions
| Expression | Substitution |
|---|---|
| \(\sqrt{{{a}^{2}}-{{x}^{2}}}\) | x = asinθ (or) acosθ |
| \(\sqrt{{{a}^{2}}+{{x}^{2}}}\) | x = atanθ (or) a cotθ |
| \(\sqrt{{{x}^{2}}-{{a}^{2}}}\) | x = asecθ (or) acosecθ |
| \(\sqrt{\frac{a+x}{a-x}}\) (or) \(\sqrt{\frac{a-x}{a+x}}\) | x = acosθ (or) acos2θ |
| \(\sqrt{(a-x)(x-b)}\) (or) \(\sqrt{\frac{a-x}{x-b}}\) | x = acos²θ + b sin²θ |
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
\(-\frac{3{{x}^{2}}}{\sqrt{1-{{x}^{6}}}}+\frac{3{{y}^{2}}}{\sqrt{1-{{y}^{3}}}}\frac{dy}{dx}=0\),
\(\frac{dy}{dx}=\frac{{{x}^{2}}}{{{y}^{2}}}\sqrt{\frac{1-{{y}^{3}}}{1-{{x}^{3}}}}\).