Question

differentiate the function.
h(x) = e^{x^{6}}+ln(x)
h(x) =

Explanation:

Step1: Use sum - rule of differentiation

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

Let \(u = x^{6}\), then \(f(x)=e^{u}\). The derivative of \(e^{u}\) with respect to \(u\) is \(e^{u}\), and the derivative of \(u = x^{6}\) with respect to \(x\) is \(6x^{5}\). By the chain - rule \(\frac{df}{dx}=\frac{df}{du}\cdot\frac{du}{dx}\), so \(f^{\prime}(x)=e^{x^{6}}\cdot6x^{5}\).

Step4: Combine the derivatives

\(h^{\prime}(x)=f^{\prime}(x)+g^{\prime}(x)=6x^{5}e^{x^{6}}+\frac{1}{x}\)

Answer:

\(6x^{5}e^{x^{6}}+\frac{1}{x}\)