Question

find dy/dx if y =√ax2 +bx +c​

Answer 1

Given :

$y = \sqrt{a} \: {x}^{2} + bx + c$

To find :

$\dfrac{dy}{dx}$

Solution :

$y = \sqrt{a} \: {x}^{2} + bx + c$

Now differentiating both sides :-

$\dfrac{dy}{dx} = \dfrac{d( \sqrt{a}{x}^{2} + bx + c)}{dx}$

$\dfrac{dy}{dx} = \dfrac{d( \sqrt{a} {x}^{2} ) }{dx} + \dfrac{d( bx )}{dx} + \dfrac{d( c )}{dx}$

$\dfrac{dy}{dx} = \sqrt{a} \: \dfrac{d( {x}^{2} ) }{dx} +b \: \dfrac{d( x )}{dx} + \dfrac{d( c )}{dx}$

$\dfrac{dy}{dx} = 2\sqrt{a} \: x\: +b + 0$

$\dfrac{dy}{dx} = 2\sqrt{a} \: x\: +b$

Answer :

$\dfrac{dy}{dx} = 2\sqrt{a} \: x\: +b$

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$\large\bf\blue{Additional-Information}$

d(sinx)/dx = cosx

d(cos x)/dx = -sin x

d(cosec x)/dx = -cot x cosec x

d(tan x)/dx = sec²x

d(sec x)/dx = secx tanx

d(cot x)/dx = - cosec² x

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Answer 2

Given: $\[y = \sqrt a {x^2} + bx + c\]$.

To find: $\[\frac{{dy}}{{dx}}\]$

Solution:

Differentiate y with respect to x.

$\[y = \sqrt a {x^2} + bx + c\]$

$\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{{d\left( {\sqrt a {x^2} + bx + c} \right)}}{{dx}}\]$

$\[ \Rightarrow \frac{{dy}}{{dx}} = \frac{{d\left( {\sqrt a {x^2}} \right)}}{{dx}} + \frac{{d\left( {bx} \right)}}{{dx}} + \frac{{d\left( c \right)}}{{dx}}\]$

$\[ \Rightarrow \frac{{dy}}{{dx}} = \sqrt a \frac{{d\left( {{x^2}} \right)}}{{dx}} + b\frac{{d\left( x \right)}}{{dx}} + c\frac{{d\left( 1 \right)}}{{dx}}\]$

$\[ \Rightarrow \frac{{dy}}{{dx}} = 2\sqrt a x + b + 0\]$

$\[ \Rightarrow \frac{{dy}}{{dx}} = 2\sqrt a + b\]$