calculate the derivative of the function ( y=(x^{4}+6)^{x} ) (i) using the fact that ( b^{x}=e^{x ln b} )…

calculate the derivative of the function ( y=(x^{4}+6)^{x} ) (i) using the fact that ( b^{x}=e^{x ln b} ) and (ii) by using logarithmic differentiation. verify that both answers are the same. (i) calculate the derivative of ( y=(x^{4}+6)^{x} ) using the fact that ( b^{x}=e^{x ln b} ). ( \frac{d y}{d x}= ) (type an exact answer. use parentheses to clearly denote the argument of each function.)

calculate the derivative of the function ( y=(x^{4}+6)^{x} ) (i) using the fact that ( b^{x}=e^{x ln b} ) and (ii) by using logarithmic differentiation. verify that both answers are the same. (i) calculate the derivative of ( y=(x^{4}+6)^{x} ) using the fact that ( b^{x}=e^{x ln b} ). ( \frac{d y}{d x}= ) (type an exact answer. use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Rewrite the function

Given (y=(x^{4}+6)^{x}), using (b^{x}=e^{x\ln b}), we rewrite (y) as (y = e^{x\ln(x^{4}+6)}).

Step2: Apply the chain rule

The chain rule states that if (y = e^{u}), then (\frac{dy}{dx}=e^{u}\cdot\frac{du}{dx}). Here (u = x\ln(x^{4}+6)). First, find (\frac{du}{dx}) using the product rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = x) and (v=\ln(x^{4}+6)). (u^\prime=1), (v^\prime=\frac{4x^{3}}{x^{4}+6}). So (\frac{du}{dx}=\ln(x^{4}+6)+x\cdot\frac{4x^{3}}{x^{4}+6}=\ln(x^{4}+6)+\frac{4x^{4}}{x^{4}+6}). Then (\frac{dy}{dx}=e^{x\ln(x^{4}+6)}\left(\ln(x^{4}+6)+\frac{4x^{4}}{x^{4}+6}\right)). Since (e^{x\ln(x^{4}+6)}=(x^{4}+6)^{x}), we have (\frac{dy}{dx}=(x^{4}+6)^{x}\left(\ln(x^{4}+6)+\frac{4x^{4}}{x^{4}+6}\right)).

Answer:

((x^{4}+6)^{x}\left(\ln(x^{4}+6)+\frac{4x^{4}}{x^{4}+6}\right))