Differentiation by Substitution Methods
Substitution is useful to reduce the function into simple form. For problems involving inverse trigonometric functions, first try for a suitable substitution to simplify it and then differentiate it. If no such substitution is found, then differentiate directly. Some standard substitutions are given below.
|
S.No |
Expressions |
Substitutions |
|
1 |
\(\sqrt{{{a}^{2}}+{{x}^{2}}}\) | x = atanθ or x = acosθ |
| 2 | \(\sqrt{{{a}^{2}}-{{x}^{2}}}\) |
x = asinθ or x = acosθ |
|
3 |
\(\sqrt{{{x}^{2}}-{{a}^{2}}}\) | x = asecθ or x = acosecθ |
| 4 | \(\frac{a+x}{a-x}\) or \(\frac{a-x}{a+x}\) |
x = atanθ |
|
5 |
\(\sqrt{\frac{a+x}{a-x}}\) or \(\sqrt{\frac{a-x}{a+x}}\) | x = acosθ |
| 6 | \(\frac{2x}{1+{{x}^{2}}}\) or \(\frac{2x}{1-{{x}^{2}}}\) |
x = atanθ |
|
7 |
acosx + bcosx | a = rcosα, b = rsinα |
| 8 | \(\sqrt{x-\alpha }\) and \(\sqrt{\beta -x}\) |
x = αsin²θ + βcos²θ |
|
9 |
\(\sqrt{2ax-{{x}^{2}}}\) |
x = a(1 – cosθ) |
Example: Find the derivative of \({{\tan }^{-1}}\left[ \frac{\sqrt{1+{{x}^{2}}}+\sqrt{1-{{x}^{2}}}}{\sqrt{1+{{x}^{2}}}-\sqrt{1-{{x}^{2}}}} \right]\) with respect to x.
Solution: Assume x² = cos2θ to simplify and then use the relation 1 + cos2θ = 2cos²θ and 1 -cos2θ = 2sin²θ to simplify the relation.
On putting x² = cos2θ we get
\(y={{\tan }^{-1}}\left[ \frac{\sqrt{1+\cos 2\theta }+\sqrt{1-\cos 2\theta }}{\sqrt{1+\cos 2\theta }-\sqrt{1-\cos 2\theta }} \right]\),
⇒ \(y\,=\,{{\tan }^{-1}}\left[ \frac{\sqrt{2{{\cos }^{2}}\theta }+\sqrt{2{{\sin }^{2}}\theta }}{\sqrt{2{{\cos }^{2}}\theta }-\sqrt{2{{\sin }^{2}}\theta }} \right]\),
⇒ \(y\,=\,{{\tan }^{-1}}\left[ \frac{\cos \theta +\sin \theta }{\cos \theta -\sin \theta } \right]\,=\,{{\tan }^{-1}}\left[ \frac{1+\tan \theta }{1-\tan \theta } \right]\),
⇒ \(y\,=\,{{\tan }^{-1}}\left[ \tan \left( \frac{\pi }{4}+\theta \right) \right]\),
⇒ \(y\,=\,\frac{\pi }{4}\,+\,\theta \),
⇒ \(y\,=\,\frac{\pi }{4}\,+\,\frac{1}{2}{{\cos }^{-1}}\,{{x}^{2}}\),
On differentiating with respect to x we get
∴ \(\frac{dy}{dx}\,=\,\frac{d}{dx}\left( \frac{\pi }{4} \right)\,+\,\frac{1}{2}\frac{d}{dx}\left( {{\cos }^{-1}}\,{{x}^{2}} \right)\),
⇒ \(\frac{dy}{dx}\,=\,0\,+\,\frac{1}{2}\frac{-1}{\sqrt{1-{{x}^{2}}}}\frac{d}{dx}\left( {{x}^{2}} \right)\),
⇒ \(\frac{dy}{dx}\,=\,-\frac{1}{2}\frac{2x}{\sqrt{1-{{x}^{4}}}}\,=\,\frac{-x}{\sqrt{1-{{x}^{4}}}}\).