10.1 Questions related to Differentiation using substitution

3.1 Scalar Differentiation:
3.2 Product Rule:
3.3 Quotient Rule:
3.4 Chain Rule

4.1 Derivative of Trigonometric functions
4.2 Derivative of Inverse Trigonometric functions
4.3 Derivative of Exponential and Logarithmic Function
4.4 Derivative Of Hyperbolic function
4.5 Questions related to derivation of common functions

8.1 Differentiation of determinant matrix

10.1 Questions related to Differentiation using substitution

Question 1: Find derivative of $\sqrt {{a^2} - {x^2}} $

$$ y = \sqrt {{a^2} - {x^2}} $$

Put,

$$ x = a\sin \theta $$$$ \frac{{dx}}{{d\theta }} = a\cos \theta $$$$ \frac{{d\theta }}{{dx}} = \frac{1}{{a\cos \theta }} $$$$ y = \sqrt {{a^2} - {{\left( {a\sin \theta } \right)}^2}} $$$$ y = \sqrt {{a^2} - {a^2}{{\sin }^2}\theta } $$$$ y = \sqrt {{a^2}\left( {1 - {{\sin }^2}\theta } \right)} $$

Using identity,

$$ {\sin ^2}\theta + {\cos ^2}\theta = 1 $$$$ y = \sqrt {{a^2}\left( {{{\cos }^2}\theta } \right)} $$$$ y = a\cos \theta $$$$ \frac{{dy}}{{d\theta }} = - a\sin \theta $$$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$$$ \frac{{dy}}{{dx}} = - a\sin \theta \left( {\frac{1}{{a\cos \theta }}} \right) = - \frac{{\sin \theta }}{{\cos \theta }} $$$$ \frac{{dy}}{{dx}} = - \tan \theta $$$$ \frac{{dy}}{{dx}} = - \frac{x}{{\sqrt {{a^2} - {x^2}} }} $$

Question 2: Find derivative of $\sqrt {{x^2} - {a^2}} $

$$ y = \sqrt {{x^2} - {a^2}} $$Put,$$ x = a\sec \theta$$$$ \frac{{dx}}{{d\theta }} = a\sec \theta \tan \theta $$$$ \frac{{d\theta }}{{dx}} = \frac{1}{{a\sec \theta \tan \theta }} $$$$ y = \sqrt {{{\left( {a\sec \theta } \right)}^2} - {a^2}} $$$$ y = \sqrt {{a^2}{{\sec }^2}\theta - {a^2}} $$$$ y = \sqrt {{a^2}\left( {{{\sec }^2}\theta - 1} \right)} $$

Using identity,

$$ {\sec ^2}\theta = 1 + {\tan ^2}\theta $$$$ y = \sqrt {{a^2}\left( {{{\tan }^2}\theta } \right)} $$$$ y = a\tan \theta $$$$ \frac{{dy}}{{d\theta }} = a{\sec ^2}\theta $$$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$$$ \frac{{dy}}{{dx}} = a{\sec ^2}\theta \left( {\frac{1}{{a\sec \theta \tan \theta }}} \right) = \frac{{\sec \theta }}{{\tan \theta }} $$$$ \frac{{dy}}{{dx}} = \frac{{\frac{1}{{\cos \theta }}}}{{\frac{{\sin \theta }}{{\cos \theta }}}} = \frac{1}{{\sin \theta }} $$$$ \frac{{dy}}{{dx}} = \cos ec\theta $$$$ \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} - {a^2}} }} $$

Question 3: Find derivative of $\sqrt {{x^2} + {a^2}} $

Solution:
$$ y = \sqrt {{x^2} + {a^2}} $$Put,$$ x = a\tan \theta $$$$ \frac{{dx}}{{d\theta }} = a{\sec ^2}\theta $$$$ \frac{{d\theta }}{{dx}} = \frac{1}{{a{{\sec }^2}\theta }} $$$$ y = \sqrt {{{\left( {a\tan \theta } \right)}^2} + {a^2}} $$$$ y = \sqrt {{a^2}{{\tan }^2}\theta + {a^2}} $$$$ y = \sqrt {{a^2}\left( {{{\tan }^2}\theta + 1} \right)} $$

Using identity,$$ {\sec ^2}\theta = 1 + {\tan ^2}\theta $$$$ y = \sqrt {{a^2}\left( {{{\sec }^2}\theta } \right)} $$$$ y = a\sec \theta $$$$ \frac{{dy}}{{d\theta }} = a\sec \theta \tan \theta $$$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$$$ \frac{{dy}}{{dx}} = a\sec \theta \tan \theta \left( {\frac{1}{{a{{\sec }^2}\theta }}} \right) = \frac{{\tan \theta }}{{\sec \theta }} $$$$ \frac{{dy}}{{dx}} = \frac{{\frac{{\sin \theta }}{{\cos \theta }}}}{{\frac{1}{{\cos \theta }}}} = \sin \theta $$$$ \frac{{dy}}{{dx}} = \sin \theta $$$$ \frac{{dy}}{{dx}} = \frac{x}{{\sqrt {{x^2} + {a^2}} }} $$

Question 4: Find derivative of $\sqrt {\frac{x}{{a + x}}} $

Solution:

$$ y = \sqrt {\frac{x}{{a + x}}} $$

Put,$$ x = a{\tan ^2}\theta $$$$ \frac{{dx}}{{d\theta }} = 2a\tan \theta {\sec ^2}\theta $$$$ \frac{{d\theta }}{{dx}} = \frac{1}{{2a\tan \theta {{\sec }^2}\theta }} = \frac{1}{{2a\tan \theta \left( {1 + {{\tan }^2}\theta } \right)}} \quad ....(1) $$

On substituting,$$ y = \sqrt {\frac{{a{{\tan }^2}\theta }}{{a + a{{\tan }^2}\theta }}} $$$$ y = \sqrt {\frac{{a{{\tan }^2}\theta }}{{a\left( {1 + {{\tan }^2}\theta } \right)}}} $$$$ y = \sqrt {\frac{{{{\tan }^2}\theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}} $$$$ \frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{1}{{\left( {\sqrt {\frac{{{{\tan }^2}\theta }}{{\left( {1 + {{\tan }^2}\theta } \right)}}} } \right)}}\left( {\frac{{\left( {1 + {{\tan }^2}\theta } \right)2\tan \theta {{\sec }^2}\theta - \left( {{{\tan }^2}\theta } \right)2\tan \theta {{\sec }^2}\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{{\sqrt {\left( {1 + {{\tan }^2}\theta } \right)} }}{{\left( {\sqrt {{{\tan }^2}\theta } } \right)}}\left( {\frac{{2\tan \theta {{\sec }^2}\theta }}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{{\sqrt {\left( {1 + {{\tan }^2}\theta } \right)} }}{{\left( {\tan \theta } \right)}}\left( {\frac{{2\tan \theta \left( {1 + {{\tan }^2}\theta } \right)}}{{{{\left( {1 + {{\tan }^2}\theta } \right)}^2}}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \left( {\frac{{\sqrt {\left( {1 + {{\tan }^2}\theta } \right)} }}{{\left( {1 + {{\tan }^2}\theta } \right)}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \frac{1}{{\sqrt {\left( {1 + {{\tan }^2}\theta } \right)} }} \quad...(2) $$

Now we know,

$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$

From $(1)$ and $(2)$,

$$ \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {\left( {1 + {{\tan }^2}\theta } \right)} }}\left( {\frac{1}{{2a\tan \theta \left( {1 + {{\tan }^2}\theta } \right)}}} \right) $$

Substitute,

$$ \frac{x}{a} = {\tan ^2}\theta $$$$ \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {\left( {1 + \frac{x}{a}} \right)} }}\left( {\frac{1}{{2a\sqrt {\frac{x}{a}} \left( {1 + \frac{x}{a}} \right)}}} \right) $$$$ \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {\left( {\frac{{a + x}}{a}} \right)} }}\left( {\frac{1}{{2a\sqrt {\frac{x}{a}} \left( {\frac{{a + x}}{a}} \right)}}} \right) $$$$ \frac{{dy}}{{dx}} = \frac{1}{{\sqrt {a + x} }}\left( {\frac{{\sqrt a \sqrt a }}{{2\sqrt x \left( {a + x} \right)}}} \right) $$$$ \frac{{dy}}{{dx}} = \frac{1}{{2\sqrt x {{\left( {a + x} \right)}^{\frac{3}{2}}}}} $$

Question 5: Find derivative of $\sqrt {\frac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}}} $

Solution:

$$ y = \frac{{{a^2} - {x^2}}}{{{a^2} + {x^2}}} $$

Substitute ${x^2} = {a^2}\cos 2\theta $,

$$ 2x\frac{{dx}}{{d\theta }} = - 2{a^2}\sin 2\theta $$$$ \frac{{dx}}{{d\theta }} = - \frac{{{a^2}\sin 2\theta }}{x} $$$$ \frac{{d\theta }}{{dx}} = - \frac{x}{{{a^2}\sin 2\theta }}\;....(1) $$

Now,$$ y = \frac{{{a^2} - {a^2}\cos 2\theta }}{{{a^2} + {a^2}\cos 2\theta }} $$$$ y = \frac{{{a^2}\left( {1 - \cos 2\theta } \right)}}{{{a^2}\left( {1 + \cos 2\theta } \right)}} = \frac{{\left( {1 - \cos 2\theta } \right)}}{{\left( {1 + \cos 2\theta } \right)}} $$

Using identity,$$ 1 - \cos 2\theta = 2{\sin ^2}\theta $$$$ 1 + \cos 2\theta = 2{\cos ^2}\theta $$

Thus,$$ y = \frac{{2{{\sin }^2}\theta }}{{2{{\cos }^2}\theta }} = {\tan ^2}\theta $$$$ \frac{{dy}}{{d\theta }} = 2\tan \theta {\sec ^2}\theta \;....(2) $$$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$

From $(1)$and $(2)$,$$ \frac{{dy}}{{dx}} = 2\tan \theta {\sec ^2}\theta \left( { - \frac{x}{{{a^2}\sin 2\theta }}} \right) $$

Using Identity,$$ \sin 2\theta = 2\sin \theta \cos \theta $$$$ \frac{{dy}}{{dx}} = 2\frac{{\sin \theta }}{{\cos \theta {{\cos }^2}\theta }}\left( { - \frac{x}{{{a^2}2\sin \theta \cos \theta }}} \right) $$$$ \frac{{dy}}{{dx}} = - \frac{x}{{{a^2}{{\cos }^4}\theta }} = - \frac{x}{{{a^2}{{\left( {{{\cos }^2}\theta } \right)}^2}}}\;...(3) $$$$ By\;{x^2} = {a^2}\cos 2\theta $$$$ \frac{{{x^2}}}{{{a^2}}} = \cos 2\theta \; = 2{\cos ^2}\theta - 1 $$$$ {\cos ^2}\theta = \frac{1}{2}\left( {\frac{{{x^2}}}{{{a^2}}} + 1} \right) $$

Substitute in $(3)$,$$ \frac{{dy}}{{dx}} = - \frac{x}{{{a^2}{{\left( {\frac{1}{2}\left( {\frac{{{x^2}}}{{{a^2}}} + 1} \right)} \right)}^2}}} = - \frac{x}{{{a^2}{{\left( {\frac{{{x^2} + {a^2}}}{{2{a^2}}}} \right)}^2}}} = - \frac{{4x{a^2}}}{{{{\left( {{x^2} + {a^2}} \right)}^2}}} $$

Question 6: Find derivative of $\sqrt {\frac{{a - x}}{{a + x}}} $

Solution: $$ y = \frac{{a - x}}{{a + x}} $$Substitute $x = a\cos 2\theta $,$$ \frac{{dx}}{{d\theta }} = - 2a\sin 2\theta $$$$ \frac{{d\theta }}{{dx}} = - \frac{1}{{2a\sin 2\theta }}\;....(1) $$Now,$$ y = \frac{{a - a\cos 2\theta }}{{a + a\cos 2\theta }} $$$$ y = \frac{{a\left( {1 - \cos 2\theta } \right)}}{{a\left( {1 + \cos 2\theta } \right)}} = \frac{{\left( {1 - \cos 2\theta } \right)}}{{\left( {1 + \cos 2\theta } \right)}} $$

Using identity, $$ 1 - \cos 2\theta = 2{\sin ^2}\theta $$$$ 1 + \cos 2\theta = 2{\cos ^2}\theta $$

Thus,$$ y = \frac{{2{{\sin }^2}\theta }}{{2{{\cos }^2}\theta }} = {\tan ^2}\theta $$$$ \frac{{dy}}{{d\theta }} = 2\tan \theta {\sec ^2}\theta \;....(2) $$$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$

From $(1)$ and $(2)$,$$ \frac{{dy}}{{dx}} = 2\tan \theta {\sec ^2}\theta \left( { - \frac{1}{{2a\sin 2\theta }}} \right) $$

$$ \sin 2\theta = 2\sin \theta \cos \theta $$$$ \frac{{dy}}{{dx}} = \frac{{\sin \theta }}{{\cos \theta {{\cos }^2}\theta }}\left( { - \frac{1}{{a2\sin \theta \cos \theta }}} \right) $$$$ \frac{{dy}}{{dx}} = - \frac{1}{{2a{{\cos }^4}\theta }} = - \frac{1}{{2a{{\left( {{{\cos }^2}\theta } \right)}^2}}}\;...(3) $$

By $x = a\cos 2\theta$, $$ \frac{x}{a} = \cos 2\theta \; = 2{\cos ^2}\theta - 1 $$$$ {\cos ^2}\theta = \frac{1}{2}\left( {\frac{x}{a} + 1} \right) $$

Substitute in $(3)$,

$$ \frac{{dy}}{{dx}} = - \frac{1}{{2a{{\left( {\frac{1}{2}\left( {\frac{x}{a} + 1} \right)} \right)}^2}}} = - \frac{1}{{2a{{\left( {\frac{{x + a}}{{2a}}} \right)}^2}}} = - \frac{{2a}}{{{{\left( {x + a} \right)}^2}}} $$

Question 7: Find derivative of $\sqrt {ax - {x^2}} $

Solution:

$$ y = \sqrt {ax - {x^2}} $$ Put,$$ x = a{\sin ^2}\theta $$$$ \frac{{dx}}{{d\theta }} = 2a\sin \theta \cos \theta $$$$ \frac{{d\theta }}{{dx}} = \frac{1}{{2a\sin \theta \cos \theta }} = \frac{1}{{a\sin 2\theta }}....(1) $$

On substituting,

$$ y = \sqrt {a\left( {a{{\sin }^2}\theta } \right) - {{\left( {a{{\sin }^2}\theta } \right)}^2}} $$$$ y = \sqrt {{a^2}{{\sin }^2}\theta - {a^2}{{\sin }^4}\theta } $$$$ y = \sqrt {{a^2}{{\sin }^2}\theta \left( {1 - {{\sin }^2}\theta } \right)} $$$$ y = \sqrt {{a^2}{{\sin }^2}\theta {{\cos }^2}\theta } $$ $$ y = a\sin \theta \cos \theta $$$$ \frac{{dy}}{{d\theta }} = a\left[ { - \sin \theta \sin \theta + \cos \theta \cos \theta } \right] $$$$ \frac{{dy}}{{d\theta }} = a\left( {{{\cos }^2}\theta - {{\sin }^2}\theta } \right) $$$$ \frac{{dy}}{{d\theta }} = a\left( {1 - {{\sin }^2}\theta - {{\sin }^2}\theta } \right) $$$$ \frac{{dy}}{{d\theta }} = a\left( {1 - 2{{\sin }^2}\theta } \right)....(2) $$

Now we know,

$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$

From $(1)$and $(2) $,

$$ \frac{{dy}}{{dx}} = a\left( {1 - 2{{\sin }^2}\theta } \right)\left( {\frac{1}{{2a\sin \theta \cos \theta }}} \right) = a\left( {1 - 2{{\sin }^2}\theta } \right)\left( {\frac{1}{{2a\sin \theta \sqrt {1 - {{\sin }^2}\theta } }}} \right) $$

Substitute,

$$ \frac{x}{a} = {\sin ^2}\theta $$$$ \frac{{dy}}{{dx}} = a\left( {1 - 2\frac{x}{a}} \right)\left( {\frac{1}{{2a\sqrt {\left( {\frac{x}{a}} \right)} \sqrt {1 - \frac{x}{a}} }}} \right) $$$$ \frac{{dy}}{{dx}} = \frac{{a\left( {a - 2x} \right)}}{a}\left( {\frac{1}{{2\sqrt a \sqrt x \sqrt {\frac{{a - x}}{a}} }}} \right) $$$$ \frac{{dy}}{{dx}} = \frac{{a - 2x}}{{2\sqrt x \sqrt {\left( {a - x} \right)} }} = \frac{{a - 2x}}{{2\sqrt {ax - {x^2}} }} $$

Question 8: Find derivative of $\sqrt {\frac{x}{{a - x}}} $

Solution:

$$ y = \sqrt {\frac{x}{{a - x}}} $$Put,$$ x = a{\sin ^2}\theta $$$$ \frac{{dx}}{{d\theta }} = 2a\sin \theta \cos \theta $$$$ \frac{{d\theta }}{{dx}} = \frac{1}{{2a\sin \theta \cos \theta }} = \frac{1}{{a\sin 2\theta }}....(1) $$

On substituting,

$$ y = \sqrt {\frac{{a{{\sin }^2}\theta }}{{a - a{{\sin }^2}\theta }}} $$$$ y = \sqrt {\frac{{a{{\sin }^2}\theta }}{{a\left( {1 - {{\sin }^2}\theta } \right)}}} $$$$ y = \sqrt {\frac{{{{\sin }^2}\theta }}{{\left( {1 - {{\sin }^2}\theta } \right)}}} $$$$ \frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{1}{{\left( {\sqrt {\frac{{{{\sin }^2}\theta }}{{\left( {1 - {{\sin }^2}\theta } \right)}}} } \right)}}\left( {\frac{{\left( {1 - {{\sin }^2}\theta } \right)2\sin \theta \cos \theta + \left( {{{\sin }^2}\theta } \right)2\sin \theta \cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{{\sqrt {\left( {1 - {{\sin }^2}\theta } \right)} }}{{\left( {\sqrt {{{\sin }^2}\theta } } \right)}}\left( {\frac{{2\sin \theta \cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \frac{1}{2}\frac{{\sqrt {\left( {1 - {{\sin }^2}\theta } \right)} }}{{\left( {\sin \theta } \right)}}\left( {\frac{{2\sin \theta \cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \left( {\frac{{\cos \theta \sqrt {\left( {1 - {{\sin }^2}\theta } \right)} }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^2}}}} \right) $$$$ \frac{{dy}}{{d\theta }} = \frac{{\cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{\frac{3}{2}}}}}....(2) $$

Now we know,

$$ \frac{{dy}}{{dx}} = \frac{{dy}}{{d\theta }}\frac{{d\theta }}{{dx}} $$

From $(1)$ and $(2)$,

$$ \frac{{dy}}{{dx}} = \frac{{\cos \theta }}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{\frac{3}{2}}}}}\left( {\frac{1}{{2a\sin \theta \cos \theta }}} \right) = \frac{1}{{{{\left( {1 - {{\sin }^2}\theta } \right)}^{\frac{3}{2}}}}}\left( {\frac{1}{{2a\sin \theta }}} \right) $$

Substitute,

$$ \frac{x}{a} = {\sin ^2}\theta $$$$ \frac{{dy}}{{dx}} = \frac{1}{{{{\left( {1 - \frac{x}{a}} \right)}^{\frac{3}{2}}}}}\left( {\frac{1}{{2a\sqrt {\frac{x}{a}} }}} \right) $$$$ \frac{{dy}}{{dx}} = \frac{1}{{{{\left( {\frac{{a - x}}{a}} \right)}^{\frac{3}{2}}}}}\left( {\frac{1}{{2\sqrt a \sqrt x }}} \right) $$$$ \frac{{dy}}{{dx}} = \frac{{a\sqrt a }}{{2\sqrt x {{\left( {a - x} \right)}^{\frac{3}{2}}}\sqrt a }} = \frac{a}{{2\sqrt x {{\left( {a - x} \right)}^{\frac{3}{2}}}}} $$