attempt 1: 2 attempts remaining. let $f(x)=x^{7x}$. use logarithmic differentiation to determine the…

Explanation:

Step1: Take natural - log of both sides

Take the natural logarithm of (y = f(x)=x^{7x}), so (\ln y=\ln(x^{7x})). Using the property of logarithms (\ln(a^{b}) = b\ln(a)), we get (\ln y = 7x\ln x).

Step2: Differentiate both sides with respect to (x)

Differentiate the left - hand side using the chain rule. The derivative of (\ln y) with respect to (x) is (\frac{1}{y}y'). Differentiate the right - hand side using the product rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = 7x) and (v=\ln x). (u^\prime = 7) and (v^\prime=\frac{1}{x}). So ((7x\ln x)^\prime=7\ln x+7x\cdot\frac{1}{x}=7\ln x + 7). Then (\frac{1}{y}y'=7\ln x + 7).

Step3: Solve for (y')

Multiply both sides by (y). Since (y = x^{7x}), we have (y'=f'(x)=x^{7x}(7\ln x + 7)).

Step4: Find (f'(1))

Substitute (x = 1) into (f'(x)). Since (\ln(1)=0), then (f'(1)=1^{7\times1}(7\ln(1)+7)=1\times(0 + 7)=7).

Answer:

(f'(x)=x^{7x}(7\ln x + 7)) (f'(1)=7)