differentiate the function.\nf(x)=2x ln(9x)-2x\nf(x)=\n\nresources\nread it\n\nsubmit answer\n\n2. - / 1…

differentiate the function.\nf(x)=2x ln(9x)-2x\nf(x)=\n\nresources\nread it\n\nsubmit answer\n\n2. - / 1 points\ndifferentiate the function.\nf(x)=sin(9 ln(x))\nf(x)=\n\nresources\nread it watch it master it

differentiate the function.\nf(x)=2x ln(9x)-2x\nf(x)=\n\nresources\nread it\n\nsubmit answer\n\n2. - / 1 points\ndifferentiate the function.\nf(x)=sin(9 ln(x))\nf(x)=\n\nresources\nread it watch it master it

Answer

Explanation:

Step1: Differentiate first - term of (f(x) = 2x\ln(9x)-2x) using product rule

The product rule ((uv)^\prime=u^\prime v + uv^\prime), where (u = 2x), (u^\prime=2), (v=\ln(9x)), (v^\prime=\frac{9}{9x}=\frac{1}{x}). So ((2x\ln(9x))^\prime=2\ln(9x)+2x\cdot\frac{1}{x}=2\ln(9x) + 2).

Step2: Differentiate second - term of (f(x))

The derivative of (-2x) is (-2).

Step3: Find (f^\prime(x))

(f^\prime(x)=(2x\ln(9x))^\prime-(2x)^\prime=2\ln(9x)+2 - 2=2\ln(9x))

Step4: Differentiate (f(x)=\sin(9\ln(x))) using chain - rule

Let (u = 9\ln(x)), then (f(x)=\sin(u)). The chain - rule states that (f^\prime(x)=\frac{df}{du}\cdot\frac{du}{dx}). (\frac{df}{du}=\cos(u)=\cos(9\ln(x))), (\frac{du}{dx}=\frac{9}{x}). So (f^\prime(x)=\frac{9\cos(9\ln(x))}{x})

Answer:

  1. (2\ln(9x))
  2. (\frac{9\cos(9\ln(x))}{x})