y=(7+e^{x})^{x}\nuse logarithmic differentiation to find \\( \\frac{d y}{d x} \\)\n\\( \\frac{d y}{d x}=…

y=(7+e^{x})^{x}\nuse logarithmic differentiation to find \\( \\frac{d y}{d x} \\)\n\\( \\frac{d y}{d x}= \\)\n\\( \\frac{x^{x}left(x^{2}+6\right)left(ln left(x^{2}+6\right)+\\frac{2 x^{2}}{x^{2}+6}\right)}{} \\)\ntype \\( \\sin (x) \\) for \\( \\sin (x), \\cos (x) \\) for \\( \\cos (x) \\), and so on.\nuse \\( x \\wedge 2 \\) to square \\( x, x \\wedge 3 \\) to cube \\( x \\), and so on.\nuse \\( (\\sin (x)) \\wedge 2 \\) to square \\( \\sin (x) \\).\nuse \\( \\ln () \\) for the natural loganthm.\ndo not simplify your answer.\nquestion help: \\( \\square \\) video \\( \\square \\) message instructor\nsubmit question jump to answer

y=(7+e^{x})^{x}\nuse logarithmic differentiation to find \\( \\frac{d y}{d x} \\)\n\\( \\frac{d y}{d x}= \\)\n\\( \\frac{x^{x}left(x^{2}+6\right)left(ln left(x^{2}+6\right)+\\frac{2 x^{2}}{x^{2}+6}\right)}{} \\)\ntype \\( \\sin (x) \\) for \\( \\sin (x), \\cos (x) \\) for \\( \\cos (x) \\), and so on.\nuse \\( x \\wedge 2 \\) to square \\( x, x \\wedge 3 \\) to cube \\( x \\), and so on.\nuse \\( (\\sin (x)) \\wedge 2 \\) to square \\( \\sin (x) \\).\nuse \\( \\ln () \\) for the natural loganthm.\ndo not simplify your answer.\nquestion help: \\( \\square \\) video \\( \\square \\) message instructor\nsubmit question jump to answer

Answer

Explanation:

Step1: Take the natural logarithm of both sides

$$\ln y=x\ln(7 + e^{x})$$

Step2: Differentiate both sides with respect to (x)

Using the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = x) and (v=\ln(7 + e^{x})). The derivative of (u=x) is (u^\prime=1). The derivative of (v = \ln(7 + e^{x})) using the chain - rule: if (y=\ln(u)) and (u = 7+e^{x}), then (y^\prime=\frac{u^\prime}{u}). Since (u^\prime=e^{x}), (v^\prime=\frac{e^{x}}{7 + e^{x}}). So (\frac{1}{y}\frac{dy}{dx}=\ln(7 + e^{x})+x\frac{e^{x}}{7 + e^{x}})

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

Multiply both sides by (y=(7 + e^{x})^{x}) $$\frac{dy}{dx}=(7 + e^{x})^{x}\left(\ln(7 + e^{x})+\frac{xe^{x}}{7 + e^{x}}\right)$$

Answer:

(\frac{dy}{dx}=(7 + e^{x})^{x}\left(\ln(7 + e^{x})+\frac{xe^{x}}{7 + e^{x}}\right))