Question

differentiate w.r.t x. 1) y=e tan-1x , 2) y=sin 5x+२​

Answer 1

Given:

$\red \bigstar \sf \: differentiate \: with \: respect \: to \: x :$

$\\ \bf \: (1) \: y = {e}^{ {tan}^{ - 1} x} \: \: \: \: \: \: \: (2)y = sin(5x + 2)$

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To find :

$\\ \red \bigstar \rm \: \frac{dy}{dx}=?$

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Solution :

$\\ \pink \bigstar \sf \: (1) \: y = {e}^{ {tan}^{ - 1}x }$

$\implies \sf \: y = {e}^{ {tan}^{ - 1} x} \\ \\ \underline{ \bf{ \bullet \: differentating \: both \: side \: wrt \: x}} \\ \\ \implies \sf \: \frac{dy}{dx} = {e}^{ {tan}^{ - 1}x } \times \frac{1}{1 + {x}^{2} } \\ \\ \\ \implies \boxed { \sf{ \frac{dy}{dx} = {e}^{ {tan}^{ - 1}x} \times \frac{1}{1 + {x}^{2} } }}$

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$\pink \bigstar \sf \: y = sin(5x + 2)$

$\\ \implies \sf \: y = sin(5x + 2) \\ \\ \underline{ \bf{ \bullet \: differentating \: both \: side \: wrt \: x}} \\ \\ \implies \sf \: \frac{dy}{dx} = \frac{d}{dx} sin(5x + 2) \\ \\ \\ \implies \sf \: \frac{dy}{dx} = cos(5x + 2) \times 5 \\ \\ \\ \implies \boxed{ \sf{ \frac{dy}{dx} = 5cos(5x + 2)}}$

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Additional information :

(1) This differential equations can be also solve by first principle of differentiation :

$\bullet\boxed{\sf{lim_{x \to 0} \: \dfrac{f(x+h)-f(x)}{h}}}$

(2) This type of equation can be easily solve by using chain rule :

Ex:

$\bf{\bullet \frac{d}{dx} sin5x = cos5x \times 5 = 5cos5x}$