Differntiability

1 Mark Questions

1. Write the derivative of sinx with respect to

COS X. Delhi 2014C

Let <math>u = \sin x</math>

On differentiating u w.r.t. x, we get

<math display="block">\frac{du}{dx} = \cos x</math>

...(i) and <math>V = \cos x</math>

On differentiating v w.r.t. x, we get

<math display="block">\frac{dv}{dx} = -\sin x</math>

...(ii)

Now,

<math display="block">\frac{du}{dv} = \frac{du}{dx} \times \frac{dx}{dv}</math> = <math>-\frac{\cos x}{\sin x}</math> [from Eqs. (i) and (ii)]

<math>\Rightarrow</math>

<math display="block">\frac{du}{dv} = -\cot x</math> (1)

1. If <math>\cos y = x \cos (a + y)</math>, where <math>\cos a \neq \pm 1</math>,

prove that <math>\frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}</math>. Foreign 2014

Given, <math>cos y = x cos (a + y)</math>

<math display="block">x = \frac{\cos y}{\cos(a+y)}</math>

On differentiating both sides w.r.t. y, we get

<math>\frac{dx}{dy} =</math> <math display="block">\cos(a+y)\frac{d}{dy}(\cos y) - \cos y\frac{d}{dy}[\cos(a+y)]</math> <math>\cos^2(a+y)</math> [by using quotient rule]

<math display="block">\Rightarrow \frac{dx}{dy} = \frac{\cos(a+y)(-\sin y) - \cos y[-\sin(a+y)]}{\cos^2(a+y)}</math> <math>\cos^2(a+y)</math> <math display="block">\frac{\cos y \sin(a+y) - \cos(a+y) \sin y}{\cos^2(a+y)}</math> <math display="block">\frac{\sin(a+y-y)}{\cos^2(a+y)}</math> <math display="block">[\because \sin A \cos B - \cos A \sin B = \sin(A - B)]</math> <math display="block">\Rightarrow \frac{dx}{dy} = \frac{\sin a}{\cos^2(a+y)}</math> <math display="block">\Rightarrow \frac{dy}{dx} = \frac{\cos^2(a+y)}{\sin a}</math> (1)

Hence proved.

1. If <math>y = \sin^{-1} \{x\sqrt{1-x} - \sqrt{x}\sqrt{1-x^2}\}</math> and <math>0 < x < 1</math>, then find <math>\frac{dy}{dx}</math>.

All India 2014C; Delhi 2010

Firstly, convert the given expression in <math>\sin^{-1}[x\sqrt{1-y^2}-y\sqrt{1-x^2}]</math> form and then put <math>x = \sin \phi</math> and <math>y = \sin \theta</math>. Now, simplify the resulting expression and differentiate it.

Given, <math>y = \sin^{-1} \left[ x \sqrt{1 - x} - \sqrt{x} \sqrt{1 - x^2} \right]</math>

Above equation can be rewritten as