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Find the derivative of the following function.

y=ln(x^(2)+8x)
Answer: 
y^(')=

Find the derivative of the following function.\newliney=ln(x2+8x) y=\ln \left(x^{2}+8 x\right) \newlineAnswer: y= y^{\prime}=

Full solution

Q. Find the derivative of the following function.\newliney=ln(x2+8x) y=\ln \left(x^{2}+8 x\right) \newlineAnswer: y= y^{\prime}=
  1. Identify function: Identify the function to differentiate.\newlineThe function given is y=ln(x2+8x)y = \ln(x^2 + 8x). We need to find the derivative of this function with respect to xx.
  2. Apply chain rule: Apply the chain rule for differentiation.\newlineThe chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. In this case, the outer function is ln(u)\ln(u) and the inner function is u(x)=x2+8xu(x) = x^2 + 8x.
  3. Differentiate outer function: Differentiate the outer function with respect to the inner function.\newlineThe derivative of ln(u)\ln(u) with respect to uu is 1u\frac{1}{u}. So, if y=ln(u)y = \ln(u), then dydu=1u\frac{dy}{du} = \frac{1}{u}.
  4. Differentiate inner function: Differentiate the inner function with respect to xx. The inner function is u(x)=x2+8xu(x) = x^2 + 8x. Using the power rule, the derivative of x2x^2 is 2x2x, and the derivative of 8x8x is 88. Therefore, dudx=2x+8\frac{du}{dx} = 2x + 8.
  5. Apply chain rule for derivative: Apply the chain rule to find the derivative of yy with respect to xx. Using the chain rule, dydx=dydududx\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}. We have already found dydu=1u\frac{dy}{du} = \frac{1}{u} and dudx=2x+8\frac{du}{dx} = 2x + 8. Now we need to substitute uu back in terms of xx to get the final derivative.
  6. Substitute and simplify: Substitute uu with x2+8xx^2 + 8x and simplify the expression.\newlineSubstituting uu with x2+8xx^2 + 8x in the expression dydu=1u\frac{dy}{du} = \frac{1}{u}, we get dydu=1x2+8x\frac{dy}{du} = \frac{1}{x^2 + 8x}. Now, multiply this by dudx=2x+8\frac{du}{dx} = 2x + 8 to get the final derivative.\newlinedydx=(1x2+8x)(2x+8)\frac{dy}{dx} = \left(\frac{1}{x^2 + 8x}\right) \cdot (2x + 8).
  7. Final derivative: Simplify the expression for the derivative.\newlinedydx=2x+8x2+8x\frac{dy}{dx} = \frac{2x + 8}{x^2 + 8x}.\newlineThis is the final derivative of the function y=ln(x2+8x)y = \ln(x^2 + 8x).

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