frac{d}{dx} left( x^3 tan^2 frac{x}{2} right) | vedclass

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(B) To find the derivative of $f(x) = x^3 	an^2 \frac{x}{2}$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u =

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$x^3 	an \frac{x}{2} \sec^2 \frac{x}{2} + 3x^2 	an^2 \frac{x}{2}$

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(B) To find the derivative of $f(x) = x^3 	an^2 \frac{x}{2}$,we use the product rule: $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$.Let $u = x^3$ and $v = 	an^2 \frac{x}{2}$.Then $\frac{du}{dx} = 3x^2$.Using the chain rule for $v$,$\frac{dv}{dx} = 2 	an \frac{x}{2} \cdot \sec^2 \frac{x}{2} \cdot \frac{1}{2} = 	an \frac{x}{2} \sec^2 \frac{x}{2}$.Now,applying the product rule:$\frac{d}{dx} \left( x^3 	an^2 \frac{x}{2} ight) = x^3 \left( 	an \frac{x}{2} \sec^2 \frac{x}{2} ight) + 	an^2 \frac{x}{2} (3x^2)$.$= x^3 	an \frac{x}{2} \sec^2 \frac{x}{2} + 3x^2 	an^2 \frac{x}{2}$.

$\frac{d}{dx} \left( x^3 \tan^2 \frac{x}{2} \right) =$

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