If a(4+x^2)=x and y-x^3=a^2,then frac{dy}{dx} | vedclass

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(D) Given $a(4+x^2)=x$,we have $a = \frac{x}{4+x^2}$.At $x=1$,$a = \frac{1}{4+1^2} = \frac{1}{5}$.Differentiating $a(4+x^2)=x$ with respect to $x$ usi

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$\frac{381}{125}$

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(D) Given $a(4+x^2)=x$,we have $a = \frac{x}{4+x^2}$.At $x=1$,$a = \frac{1}{4+1^2} = \frac{1}{5}$.Differentiating $a(4+x^2)=x$ with respect to $x$ using the product rule:$\frac{da}{dx}(4+x^2) + a(2x) = 1$.Substituting $x=1$ and $a=\frac{1}{5}$:$\frac{da}{dx}(4+1) + \frac{1}{5}(2(1)) = 1 \implies 5\frac{da}{dx} + \frac{2}{5} = 1 \implies 5\frac{da}{dx} = \frac{3}{5} \implies \frac{da}{dx} = \frac{3}{25}$.Given $y = x^3 + a^2$,differentiating with respect to $x$:$\frac{dy}{dx} = 3x^2 + 2a\frac{da}{dx}$.At $x=1$,$a=\frac{1}{5}$ and $\frac{da}{dx}=\frac{3}{25}$:$\frac{dy}{dx} = 3(1)^2 + 2(\frac{1}{5})(\frac{3}{25}) = 3 + \frac{6}{125} = \frac{375+6}{125} = \frac{381}{125}$.

If $a(4+x^2)=x$ and $y-x^3=a^2$,then $\frac{dy}{dx}$ at $x=1$ is ...

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