Question

Please solve..
Method of differentiation

Answer 1

Question :

$\sf y = \tan {}^{ - 1} ( \frac{4x}{1 + 5x {}^{2} } ) + \tan {}^{ - 1} ( \frac{2 + 3x}{3 - 2x})$

Find $\sf\:\dfrac{dy}{dx}$

Formula's used;

•Inverse Trignometric Formulas:

$1) \sf\tan {}^{ - 1} x + \tan {}^{ - 1} y = \tan {}^{ - 1} ( \frac{x + y}{1 - xy} )$

$2) \sf \tan {}^{ - 1}x - \tan {}^{ - 1} y = \tan {}^{ - 1} ( \frac{x - y}{1 + xy} )$

• Differentiation Formula's :

$1) \dfrac{d(x {}^{n}) }{dx} = nx {}^{n - 1}$

$2) \dfrac{d(constant)}{dx} = 0$

$3) \frac{d( \tan {}^{ - 1} x)}{dx} = \frac{1}{1 + x {}^{2} }$

Solution :

$\sf y = \tan {}^{ - 1} ( \frac{4x}{1 + 5x { }^{2} } ) + \tan {}^{ - 1} ( \frac{2 + 3x}{3 - 2x})$

$\sf y = \tan {}^{ - 1} ( \frac{5x - x}{1 +( 5x )\times x} ) + \tan {}^{ - 1} ( \frac{ \frac{2}{3} + x }{1 - \frac{2}{3} \times x } )$

Now expand the terms

$\sf y = \tan{}^{ - 1}5x - \tan {}^{ - 1} x + \tan {}^{ - 1} ( \frac{2}{3}) + \tan {}^{ - 1} x$

$\sf y = \tan {}^{ - 1} 5x \cancel{ - \tan {}^{ - 1} x} + \tan {}^{ - 1} ( \frac{2}{3}) + \cancel{ \tan {}^{ - 1} x}$

$\sf y = \tan {}^{ - 1} 5x + \tan {}^{ - 1} ( \frac{2}{3} )$

Now Differentiate with respect to x

$\sf \implies \dfrac{dy}{dx} = \dfrac{d( \tan {}^{ - 1} 5x)}{d(5x)} \times \dfrac{d(5x)}{dx}$

$\sf \implies \dfrac{dy}{dx} = \dfrac{1}{1 + 25x {}^{2} } \times 5$

It is the required solution

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More About Differention:

•Chain Rule :

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$