compute the derivative. use logarithmic differentiation where appropriate.\n\n$$\\frac { d } { d x } \\left(…

compute the derivative. use logarithmic differentiation where appropriate.\n\n$$\\frac { d } { d x } \\left( 1 + \\frac { 3 } { x } \\right) ^ { x }$$\n\n$$\\frac { d } { d x } \\left( 1 + \\frac { 3 } { x } \\right) ^ { x } = \\square$$\n(use parentheses to clearly denote the argument of each function.)

compute the derivative. use logarithmic differentiation where appropriate.\n\n$$\\frac { d } { d x } \\left( 1 + \\frac { 3 } { x } \\right) ^ { x }$$\n\n$$\\frac { d } { d x } \\left( 1 + \\frac { 3 } { x } \\right) ^ { x } = \\square$$\n(use parentheses to clearly denote the argument of each function.)

Answer

Explanation:

Step1: Let ( y=(1 + \frac{3}{x})^{x} )

Take the natural logarithm of both sides: ( \ln y=x\ln(1+\frac{3}{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(1+\frac{3}{x}))

  • For (u = x), (u^\prime=1)
  • For (v=\ln(1+\frac{3}{x})), first rewrite (1+\frac{3}{x}=\frac{x + 3}{x}), then (v=\ln(x + 3)-\ln x) (v^\prime=\frac{1}{x + 3}-\frac{1}{x}=\frac{x-(x + 3)}{x(x + 3)}=-\frac{3}{x(x + 3)})

So (\frac{1}{y}y^\prime=\ln(1+\frac{3}{x})+x(-\frac{3}{x(x + 3)})=\ln(1+\frac{3}{x})-\frac{3}{x + 3})

Step3: Solve for ( y^\prime )

Since (y=(1+\frac{3}{x})^{x}), then (y^\prime=(1+\frac{3}{x})^{x}\left(\ln(1+\frac{3}{x})-\frac{3}{x + 3}\right))

Answer:

(\left(1+\frac{3}{x}\right)^{x}\left(\ln\left(1+\frac{3}{x}\right)-\frac{3}{x + 3}\right))