Question

1. - / 1 points differentiate the function. h(x)=e^{x^{3}}+ln(x) h(x)=

Explanation:

Step1: Use sum - rule of differentiation

The sum - rule states that if $h(x)=f(x)+g(x)$, then $h^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)$. Here, $f(x)=e^{x^{3}}$ and $g(x)=\ln(x)$.

Step2: Differentiate $g(x)=\ln(x)$

The derivative of $\ln(x)$ with respect to $x$ is $\frac{1}{x}$, so $g^{\prime}(x)=\frac{1}{x}$.

Step3: Differentiate $f(x)=e^{x^{3}}$ using chain - rule

Let $u = x^{3}$, then $f(x)=e^{u}$. The chain - rule states that $\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}$. The derivative of $e^{u}$ with respect to $u$ is $e^{u}$, and the derivative of $u = x^{3}$ with respect to $x$ is $3x^{2}$. So $f^{\prime}(x)=e^{u}\cdot3x^{2}=e^{x^{3}}\cdot3x^{2}$.

Step4: Combine the derivatives

$h^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)=3x^{2}e^{x^{3}}+\frac{1}{x}$.

Answer:

$3x^{2}e^{x^{3}}+\frac{1}{x}$