Question

differentiate the function. h(x)=e^{x^{6}}+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, $f(x)=e^{x^{6}}$ 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 $f(x)=e^{x^{6}}$ using chain - rule

The chain - rule states that if $y = e^{u}$ and $u = x^{6}$, then $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. The derivative of $y = e^{u}$ with respect to $u$ is $e^{u}$, and the derivative of $u=x^{6}$ with respect to $x$ is $6x^{5}$. So, $f^{\prime}(x)=e^{x^{6}}\cdot6x^{5}$.

Step4: Find $h^{\prime}(x)$

By the sum - rule, $h^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)=6x^{5}e^{x^{6}}+\frac{1}{x}$.

Answer:

$6x^{5}e^{x^{6}}+\frac{1}{x}$