write the composite function in the form f(g(x)). y = e^7√x. u = g(x) = blank y = f(u) = blank then find the…

Explanation:

Step1: Identify inner - outer functions for $y = e^{7\sqrt{x}}$

Let $u = g(x)=7\sqrt{x}=7x^{\frac{1}{2}}$, and $y = f(u)=e^{u}$.

Step2: Find derivatives of $g(x)$ and $f(u)$

The derivative of $g(x)$ using the power rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$ is $g'(x)=\frac{7}{2}x^{-\frac{1}{2}}$. The derivative of $f(u)$ is $f'(u)=e^{u}$.

Step3: Apply chain - rule $\frac{dy}{dx}=f'(g(x))\cdot g'(x)$

Substitute $u = g(x)$ into $f'(u)$ and multiply by $g'(x)$. We get $\frac{dy}{dx}=e^{7\sqrt{x}}\cdot\frac{7}{2\sqrt{x}}$.

Step4: For $f(x)=(1 + x^{3})^{\frac{3}{4}}$

Let $u = g(x)=1 + x^{3}$, and $y = f(u)=u^{\frac{3}{4}}$.

Step5: Find derivatives of $g(x)$ and $f(u)$

$g'(x)=3x^{2}$, and $f'(u)=\frac{3}{4}u^{-\frac{1}{4}}$ using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$.

Step6: Apply chain - rule $\frac{dy}{dx}=f'(g(x))\cdot g'(x)$

Substitute $u = g(x)$ into $f'(u)$ and multiply by $g'(x)$. We have $f'(x)=\frac{3}{4}(1 + x^{3})^{-\frac{1}{4}}\cdot3x^{2}=\frac{9x^{2}}{4(1 + x^{3})^{\frac{1}{4}}}$.

Answer:

For $y = e^{7\sqrt{x}}$, $u = g(x)=7\sqrt{x}$, $y = f(u)=e^{u}$, $\frac{dy}{dx}=\frac{7e^{7\sqrt{x}}}{2\sqrt{x}}$ For $f(x)=(1 + x^{3})^{\frac{3}{4}}$, $f'(x)=\frac{9x^{2}}{4(1 + x^{3})^{\frac{1}{4}}}$