7. use logarithmic differentiation to find the derivative \\( \\frac { d y } { d x } \\) for\n\n\\( y =…

Explanation:

Step1: Take natural logarithm on both sides

$$\ln y=\ln\left(\frac{x^{x^{3}}(x^{2} + 1)^{5}}{e^{x}\sqrt[3]{x + 1}}\right)$$ Using the properties of logarithms (\ln\frac{a}{b}=\ln a-\ln b), (\ln(ab)=\ln a+\ln b) and (\ln a^{b}=b\ln a), we get: $$\ln y=x^{3}\ln x + 5\ln(x^{2}+1)-x-\frac{1}{3}\ln(x + 1)$$

Step2: Differentiate both sides with respect to (x)

Differentiate term - by - term:

• For (u = x^{3}\ln x), using the product rule ((uv)^\prime=u^\prime v+uv^\prime) where (u = x^{3}), (u^\prime=3x^{2}) and (v=\ln x), (v^\prime=\frac{1}{x}). So ((x^{3}\ln x)^\prime=3x^{2}\ln x+x^{2})
• For (u = 5\ln(x^{2}+1)), using the chain rule ((\ln g(x))^\prime=\frac{g^\prime(x)}{g(x)}), (g(x)=x^{2}+1), (g^\prime(x) = 2x). So ((5\ln(x^{2}+1))^\prime=\frac{10x}{x^{2}+1})
• ((-x)^\prime=-1)
• For (u=-\frac{1}{3}\ln(x + 1)), using the chain rule, ((-\frac{1}{3}\ln(x + 1))^\prime=-\frac{1}{3(x + 1)})

So (\frac{1}{y}\frac{dy}{dx}=3x^{2}\ln x+x^{2}+\frac{10x}{x^{2}+1}-1-\frac{1}{3(x + 1)})

Step3: Solve for (\frac{dy}{dx})

Multiply both sides by (y): Since (y=\frac{x^{x^{3}}(x^{2}+1)^{5}}{e^{x}\sqrt[3]{x + 1}}) $$\frac{dy}{dx}=\frac{x^{x^{3}}(x^{2}+1)^{5}}{e^{x}\sqrt[3]{x + 1}}\left(3x^{2}\ln x+x^{2}+\frac{10x}{x^{2}+1}-1-\frac{1}{3(x + 1)}\right)$$

Answer:

$$\frac{dy}{dx}=\frac{x^{x^{3}}(x^{2}+1)^{5}}{e^{x}\sqrt[3]{x + 1}}\left(3x^{2}\ln x+x^{2}+\frac{10x}{x^{2}+1}-1-\frac{1}{3(x + 1)}\right)$$