If u and v are functions of x, then int frac{ | vedclass

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Question

(D) Consider the derivative of the quotient $\frac{u^2}{2v^2}$ with respect to $x$ using the quotient rule:$\frac{d}{dx} (\frac{u^2}{2v^2}) = \frac{1}

Vedclass · Accepted Answer

$\frac{u^2}{2v^2} + c$

Vedclass · Answer

(D) Consider the derivative of the quotient $\frac{u^2}{2v^2}$ with respect to $x$ using the quotient rule:$\frac{d}{dx} (\frac{u^2}{2v^2}) = \frac{1}{2} \cdot \frac{v^2 \frac{d}{dx}(u^2) - u^2 \frac{d}{dx}(v^2)}{(v^2)^2}$$= \frac{1}{2} \cdot \frac{v^2 (2u \frac{du}{dx}) - u^2 (2v \frac{dv}{dx})}{v^4}$$= \frac{v^2 u \frac{du}{dx} - u^2 v \frac{dv}{dx}}{v^4}$$= \frac{v(uv \frac{du}{dx} - u^2 \frac{dv}{dx})}{v^4}$$= \frac{uv \frac{du}{dx} - u^2 \frac{dv}{dx}}{v^3}$Since the derivative of $\frac{u^2}{2v^2}$ is the integrand, the integral is $\frac{u^2}{2v^2} + c$.

If $u$ and $v$ are functions of $x$, then $\int \frac{1}{v^3} (uv \frac{du}{dx} - u^2 \frac{dv}{dx}) dx =$

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