Question

Differentiate WRT x. log(x^3+2)

Answer 1

$log( {x}^{3} + 2 )$

We have to differentiate the above expression with respect to x.

$\implies \dfrac{d(log( {x}^{3} + 2 ) )}{dx}$

Now differentiate using Chain rule :-

[ d(logx)/dx = 1/x]

$\implies \dfrac{1}{({x}^{3} + 2 )} \times \dfrac{d({x}^{3} + 2 ) }{dx}$

$\implies \dfrac{1}{({x}^{3} + 2 )} \times ( \dfrac{d({x}^{3} ) }{dx} + \dfrac{d( 2 ) }{dx} )$

[d(x^n)/dx = n x^(n-1) ]

differentiation of constant term is zero. Therefore d(2)/dx is zero.

$\implies \dfrac{1}{({x}^{3} + 2 )} \times (3{x}^{2} + 0)$

$\implies \dfrac{1}{({x}^{3} + 2 )} \times 3{x}^{2}$

$\implies \dfrac{3{x}^{2}}{{x}^{3} + 2 }$

Answer :

$\dfrac{3{x}^{2}}{{x}^{3} + 2 }$

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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 ,

The function is y = log(x³ + 2)

Differentiaing y wrt x by using chain rule , we get

$\tt \implies \frac{dy}{dx} = \frac{d \{ log( {x}^{3} + 2) \}}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{ {x}^{3} + 2 } \frac{d \{ {x}^{3} + 2 \}}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{1}{ {x}^{3} + 2} \times 3 {x}^{2} + 0$

$\tt \implies\frac{dy}{dx} = \frac{ 3 {x}^{2} }{ {x}^{3} + 2}$

Remmember :

$\tt \implies\frac{d \{log (x ) \}}{dx} = \frac{1}{x}$

$\tt \implies\frac{d \{ {x}^{n} \}}{dx} =n {(x)}^{n - 1}$

$\tt \implies\frac{d(constant)}{dx} =0$

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