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I need some help finding the derivative of this function using logarithmic differentiation: $$y = \dfrac{(x + 3)^4(2x^2 + 5x)^3}{\sqrt{4x - 3}}$$

I've tried asking the teacher for help but I'm still confused on using logarithmic differentiation.

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2 Answers

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From here, $$y' = y\cdot(\ln y)'$$

These log rules will be helpful to you: $$\begin{align} \ln(ab) &= \ln a + \ln b \\ \ln\left(\dfrac ab\right) &= \ln a - \ln b \\ \ln a^b &= b\ln a \end{align}$$

And $$\dfrac{\mathrm d}{\mathrm dx}\ln\left(g(x)\right) = \dfrac{g'(x)}{g(x)}$$

Here is an example of derivating a function using the rules mentioned above. $$ \begin{align} y &= (1 + x)\left(1 + x^2\right)^2\left(1 + x^3\right)^3 \\ \ln y &= \ln \left[(1 + x)\left(1 + x^2\right)^2\left(1 + x^3\right)^3\right] \\ &= \ln(1 + x) + 2\ln\left(1 + x^2\right) + 3\ln\left(1 + x^3\right) \\ (\ln y)' &= \dfrac{1}{1 + x} + \dfrac{4x}{1 + x^2} + \dfrac{9x^2}{1 + x^3} \\ y' &= y\cdot(\ln y)' = (1 + x)\left(1 + x^2\right)^2\left(1 + x^3\right)^3\left[\dfrac{1}{1 + x} + \dfrac{4x}{1 + x^2} + \dfrac{9x^2}{1 + x^3}\right] \end{align}$$ Use this example as a guideline to solve your problem.

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The way I was taught this subject is quite simple: the logarithmic derivative of a function $f$ is the ordinary derivative of $\ln|f|$, i.e. $\frac{f'}{f}$, and it obeys the same rules as computations with logs.

Thus here, you have $\;f(x)=\dfrac{(x + 3)^4(2x^2 + 5x)^3}{\sqrt{4x - 3}}=\dfrac{(x + 3)^4 x^3(2x + 5)^3}{(4x - 3)^{\tfrac12}},\;$ so rac12$$\frac{f'}f=4\cdot\frac1{x+3}+3\cdot \frac1x+3\cdot\frac2{2x+5}-\frac12\frac4{4x-3}=\dotsm$$

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