Calculate the second derivative of the function

$$f(x)=\mathrm{ln}(4{x}^{2}-x)$$

$$f(x)=\mathrm{ln}(4{x}^{2}-x)$$

Marilyn Cameron
2022-10-16
Answered

Calculate the second derivative of the function

$$f(x)=\mathrm{ln}(4{x}^{2}-x)$$

$$f(x)=\mathrm{ln}(4{x}^{2}-x)$$

You can still ask an expert for help

Ostrakodec3

Answered 2022-10-17
Author has **18** answers

Solution:

$$f(x)=\mathrm{ln}(4{x}^{2}-x)\phantom{\rule{0ex}{0ex}}{f}^{\prime}(x)=\frac{1}{4{x}^{2}-x}(8x-1)\phantom{\rule{0ex}{0ex}}{f}^{\prime}(x)=\frac{8x-1}{4{x}^{2}-x}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=\frac{(4{x}^{2}-x)(8-0)-(8x-1)(8x-1)}{(4{x}^{2}-x{)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=\frac{8(4{x}^{2}-x)-(8x-1{)}^{2}}{(4{x}^{2}-x{)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=\frac{8x-32{x}^{2}-1}{(4{x}^{2}-x{)}^{2}}$$

$$f(x)=\mathrm{ln}(4{x}^{2}-x)\phantom{\rule{0ex}{0ex}}{f}^{\prime}(x)=\frac{1}{4{x}^{2}-x}(8x-1)\phantom{\rule{0ex}{0ex}}{f}^{\prime}(x)=\frac{8x-1}{4{x}^{2}-x}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=\frac{(4{x}^{2}-x)(8-0)-(8x-1)(8x-1)}{(4{x}^{2}-x{)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=\frac{8(4{x}^{2}-x)-(8x-1{)}^{2}}{(4{x}^{2}-x{)}^{2}}\phantom{\rule{0ex}{0ex}}{f}^{\u2033}(x)=\frac{8x-32{x}^{2}-1}{(4{x}^{2}-x{)}^{2}}$$

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