# Differentiation by Substitution Methods

## 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

∴ $$\frac{dy}{dx}\,=\,\frac{d}{dx}\left( \frac{\pi }{4} \right)\,+\,\frac{1}{2}\frac{d}{dx}\left( {{\cos }^{-1}}\,{{x}^{2}} \right)$$,

⇒ $$\frac{dy}{dx}\,=\,0\,+\,\frac{1}{2}\frac{-1}{\sqrt{1-{{x}^{2}}}}\frac{d}{dx}\left( {{x}^{2}} \right)$$,

⇒ $$\frac{dy}{dx}\,=\,-\frac{1}{2}\frac{2x}{\sqrt{1-{{x}^{4}}}}\,=\,\frac{-x}{\sqrt{1-{{x}^{4}}}}$$.