Question

can anyone please give me explanation of que. 4

Answer 1

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Here is Your Answer....!!!!
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Actually welcome to the concept of the Calculus ....

Answer 2

Given,

$y=\frac{x^{3}+4}{x+1}\\\\\text{Diff. both sides  with respect  to x}\\\\\frac{dy}{dx}=\frac{d(\frac{x^{3}+4}{x+1})}{dx}\\\\\frac{dy}{dx}=\frac{(x+1)\frac{d(x^{3}+4)}{dx}-(x^{3}+4)\frac{d(x+1)}{dx}}{(x+1)^{2}}\\\\\frac{dy}{dx}=\frac{(x+1)(3x^{2}+0)-(x^{3}+4)(1+0)}{(x+1)^{2}}\\\\\frac{dy}{dx}=\frac{3x^{3}+3x^{2}-x^{3}-4}{(x+1)^{2}}\\\\\\\frac{dy}{dx}=\frac{2x^{3}+3x^{2}-4}{(x+1)^{2}}\;\;\textbf{Ans.}\\\\\\Note;\\\frac{d(u/v)}{dx}=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^{2} }$