Question

Differentiate the following. y=[√2+x/2-x ]×(x^2+4)​

Answer 1

$\bf \implies \: y = \bigg( \sqrt{2} + \dfrac{x}{2} - x \bigg) \bigg( {x}^{2} + 4 \bigg)$

We have to differentiate the above expression.

$\bf \implies \: \dfrac{dy}{dx} = \dfrac{d \bigg[\bigg( \sqrt{2} + \dfrac{x}{2} - x \bigg) \bigg( {x}^{2} + 4 \bigg) \bigg]}{dx}$

Now we will use product rule :-

$\boxed{\large \sf\dfrac{d(uv)}{dx} = u \dfrac{dv}{dx} + v u \dfrac{du}{dx}}$

$\bf \implies \: \dfrac{dy}{dx} =({x}^{2} + 4 ) \dfrac{d\bigg( \sqrt{2} + \dfrac{x}{2} - x \bigg) }{dx} + \dfrac{d \bigg( {x}^{2} + 4 \bigg) }{dx} \bigg(\sqrt{2} + \dfrac{x}{2} - x \bigg)$

We know that :-

$\boxed{\large \sf\dfrac{d(c)}{dx} = 0 } \sf where \: c \: is \: constant$

$\boxed{\large \sf\dfrac{d( {x}^{n} )}{dx} = n {x}^{n - 1} }$

$\bf \implies \: \dfrac{dy}{dx} =({x}^{2} + 4 )\bigg( 0 + \dfrac{1}{2} - 1 \bigg) + \bigg(2 {x} + 0\bigg) \bigg(\sqrt{2} + \dfrac{x}{2} - x \bigg)$

$\bf \implies \: \dfrac{dy}{dx} =({x}^{2} + 4 )\bigg( - \dfrac{1}{2} \bigg) + \bigg(2 {x} \bigg) \bigg(\sqrt{2} - \dfrac{x}{2} \bigg)$

$\bf \implies \: \dfrac{dy}{dx} =\bigg( - \dfrac{ {x}^{2} }{2} - \dfrac{4}{2} \bigg ) + \bigg(2\sqrt{2} - {x} \bigg)$

$\bf \implies \: \dfrac{dy}{dx} = - \dfrac{ {x}^{2} }{2} - {2} + 2\sqrt{2} - {x}$

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Learn more :

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