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