Question

find the derivative of the function.
$f(x)=(5x^{6}+4x^{3})^{4}$
$f(x)=$
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find the derivative of the function.
$f(x)=sqrt{9x + 8}$
$f(x)=$
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Explanation:

Step1: Identify outer - inner functions for $F(x)$

Let $u = 5x^{6}+4x^{3}$, then $F(x)=u^{4}$.

Step2: Differentiate outer function

Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, $\frac{dF}{du}=4u^{3}=4(5x^{6}+4x^{3})^{3}$.

Step3: Differentiate inner function

$\frac{du}{dx}=\frac{d}{dx}(5x^{6}+4x^{3})=30x^{5}+12x^{2}$.

Step4: Apply chain - rule

By the chain - rule $\frac{dF}{dx}=\frac{dF}{du}\cdot\frac{du}{dx}$, so $F'(x)=4(5x^{6}+4x^{3})^{3}(30x^{5}+12x^{2})$.

Step5: Identify outer - inner functions for $f(x)$

Let $u = 9x + 8$, then $f(x)=\sqrt{u}=u^{\frac{1}{2}}$.

Step6: Differentiate outer function

Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, $\frac{df}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}=\frac{1}{2\sqrt{9x + 8}}$.

Step7: Differentiate inner function

$\frac{du}{dx}=\frac{d}{dx}(9x + 8)=9$.

Step8: Apply chain - rule

By the chain - rule $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$, so $f'(x)=\frac{9}{2\sqrt{9x + 8}}$.

Answer:

$F'(x)=4(5x^{6}+4x^{3})^{3}(30x^{5}+12x^{2})$
$f'(x)=\frac{9}{2\sqrt{9x + 8}}$