If f(x) = -(sin^2 x + cos^5 x),then find lim_ | vedclass

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(C) Given $f(x) = -(\sin^2 x + \cos^5 x)$.To find: $\lim_{x \rightarrow 0} \frac{f'(x)}{x}$.Differentiating $f(x)$ with respect to $x$ using the chain

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exists and is equal to $3$

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(C) Given $f(x) = -(\sin^2 x + \cos^5 x)$.To find: $\lim_{x ightarrow 0} \frac{f'(x)}{x}$.Differentiating $f(x)$ with respect to $x$ using the chain rule:$f'(x) = -[2 \sin x \cos x + 5 \cos^4 x(-\sin x)]$$f'(x) = -\sin x (2 \cos x - 5 \cos^4 x)$Now,evaluate the limit:$\lim_{x ightarrow 0} \frac{f'(x)}{x} = \lim_{x ightarrow 0} \frac{-\sin x (2 \cos x - 5 \cos^4 x)}{x}$$= -(\lim_{x ightarrow 0} \frac{\sin x}{x}) 	imes (\lim_{x ightarrow 0} (2 \cos x - 5 \cos^4 x))$$= -1 	imes (2(1) - 5(1)^4)$$= -1 	imes (2 - 5) = -1 	imes (-3) = 3$.Thus,the limit exists and is equal to $3$.

If $f(x) = -(\sin^2 x + \cos^5 x)$,then find $\lim_{x \rightarrow 0} \frac{f'(x)}{x}$.

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