Question

hw 12 - higher order derivatives section 2. (1 point)
let $f(x)=(3x + 29)^6$.
$f(x)=square$
$f(x)=square$
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Explanation:

Step1: Apply chain - rule for first derivative

Let $u = 3x+29$, then $f(x)=u^{6}$. The derivative of $y = u^{6}$ with respect to $u$ is $\frac{dy}{du}=6u^{5}$, and the derivative of $u = 3x + 29$ with respect to $x$ is $\frac{du}{dx}=3$. By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$, so $f^{\prime}(x)=6(3x + 29)^{5}\cdot3=18(3x + 29)^{5}$.

Step2: Apply chain - rule for second derivative

Let $v=(3x + 29)^{5}$. Then $f^{\prime}(x)=18v$. The derivative of $y = 18v$ with respect to $v$ is $\frac{dy}{dv}=18$, and the derivative of $v=(3x + 29)^{5}$ with respect to $x$ using the chain - rule: let $w = 3x+29$, $v = w^{5}$, $\frac{dv}{dw}=5w^{4}$, $\frac{dw}{dx}=3$, so $\frac{dv}{dx}=5(3x + 29)^{4}\cdot3 = 15(3x + 29)^{4}$. Then $f^{\prime\prime}(x)=18\cdot15(3x + 29)^{4}=270(3x + 29)^{4}$.

Answer:

$f^{\prime}(x)=18(3x + 29)^{5}$
$f^{\prime\prime}(x)=270(3x + 29)^{4}$