Question

differentiate x under root 1+x³ at x=2​

Answer 1

Hey There

Here's The Answer

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• Kindly refer to the Attachment

Hope It Helps.

Answer 2

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

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

Now using product rule :-

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

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

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

$\implies \: {(1 + {x}^{3} )}^{ \frac{1}{2} } + \dfrac{1}{2} x {(1 + {x}^{3} )}^{ \frac{ - 1}{2} } 3{x}^{2}$

$\implies \: {(1 + {x}^{3} )}^{ \frac{1}{2} } + \dfrac{3}{2} {(1 + {x}^{3} )}^{ \frac{ - 1}{2} } {x}^{3}$

Now to get final answer put x = 2

$\implies \: {(1 + {2}^{3} )}^{ \frac{1}{2} } + \dfrac{3}{2} {(1 + {2}^{3} )}^{ \frac{ - 1}{2} } {2}^{3}$

We know that :- 2³=8

$\implies \: {(1 + 8)}^{ \frac{1}{2} } + {3} {(1 + 8 )}^{ \frac{ - 1}{2} } {2}^{2}$

$\implies \: {(9)}^{ \frac{1}{2} } + {3} {(9 )}^{ \frac{ - 1}{2} } {2}^{2}$

$\implies \: {(3)}^{ \frac{1}{2} \times 2 } + {3} {(3 )}^{ \frac{ - 1}{2} \times 2} \: {2}^{2}$

$\implies \: {(3)} + {3} {(3 )}^{ { - 1}} \: {2}^{2}$

$\implies \: {(3)} + {(3 )}^{ { - 1 + 1}} \: {2}^{2}$

$\implies \: {(3)} + {(3 )}^{ {0}} \: {2}^{2}$

$\implies \: {3} + \: {2}^{2}$

we know that :- 2²= 4

$\implies \: {3} + 4$

$\implies \: 7$

Answer : 7