Question

exercises 3.6 the chain
score: 6/19 answered: 6/19
question 7
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if ( f(x)=sin ^{5} x ), find ( f^{prime}(x) )
find ( f^{prime}(3) )
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Explanation:

Step1: Apply the chain rule

Let $u = \sin x$, then $f(x)=u^{5}$. The derivative of $y = u^{n}$ with respect to $u$ is $y^{\prime}=nu^{n - 1}$, and the derivative of $u=\sin x$ with respect to $x$ is $u^{\prime}=\cos x$. By the chain rule $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$.
For $f(u) = u^{5}$, $\frac{df}{du}=5u^{4}$. Substituting $u = \sin x$ back in, we get $\frac{df}{du}=5\sin^{4}x$. And $\frac{du}{dx}=\cos x$.

Step2: Calculate $f^{\prime}(x)$

Using the chain - rule formula $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$, we have $f^{\prime}(x)=5\sin^{4}x\cdot\cos x$.

Step3: Calculate $f^{\prime}(3)$

Substitute $x = 3$ into $f^{\prime}(x)$. So $f^{\prime}(3)=5\sin^{4}(3)\cos(3)$.

Answer:

$f^{\prime}(x) = 5\sin^{4}x\cos x$; $f^{\prime}(3) = 5\sin^{4}(3)\cos(3)$