derivative of logs

[samantha0417]

the problem: [e^(-2X)*(2-x^3)^(3/2)] / [ sqrt of (1+ x^2)]

I'm not sure if I did this right but I got the following answer. Is this correct, if not some guidance would be appreciated

[sqrt(1+X^2) * (e^(-2X)*(9/2)(2-x^3)^(.5) + (-2e^(-2X) * (2-x^3)^(1.5)) - (-x(1+x)^(-1.5) * e^(-2X)(2-x^3)^(1.5))] / [ (sqrt of (1+ x^2))^2)]

 

[soroban]

Hello, Samantha!

I assume you're expected to use logarithmic differentiation.
. . It's very messy . . .

Differentiate: \(\displaystyle \L\:y\;=\;\frac{e^{-2x}(2\,-\,x^3)^{\frac{3}{2}}}{\sqrt{1\,+\,x^2}}\)

Take logs:
. . \(\displaystyle \L\ln(y)\;=\;\ln\left[\frac{e^{-2x}(2\,-\,x^3)^{\frac{3}{2}}}{\sqrt{1\,+\,x^2}}\right]\)

. . \(\displaystyle \L\ln(y) \;= \;\ln\left[e^{-2x}(2\,-\,x^3)^{\frac{3}{2}}\right]\,-\,\ln\left[\sqrt{1\,+\,x^2}\right]\)

. . \(\displaystyle \L\ln(y) \;= \;\ln\left[e^{-2x}\right]\,+\,\left(2\,-\,x^3\right)^{\frac{3}{2}} \,- \,\ln\left(1\,+\,x^2\right)^{\frac{1}{2}}\)

. . \(\displaystyle \L\ln(y)\;=\;-2x\cdot\ln(e)\,+\,\frac{3}{2}\cdot\ln(2\,-\,x^3)\,-\,\frac{1}{2}\cdot\ln(1\,+\,x^2)\)

. . \(\displaystyle \L\ln(y)\;=\;-2x\,+\,\frac{3}{2}\cdot\ln(2\,-\,x^3)\,-\,\frac{1}{2}\cdot\ln(1\,+\,x^2)\)


Differentiate implicitly:

. . \(\displaystyle \L\frac{1}{y}\cdot y' \;= \;-2 \,+\,\frac{3}{2}\cdot\frac{1}{2\,-\,x^3}(-3x^2) \,- \,\frac{1}{2}\cdot\frac{1}{1\,+\,x^2}(2x)\)

. . . . \(\displaystyle \L\frac{y'}{y}\;=\;-2\,-\,\frac{9x^2}{2(2\,-\,x^3)} \,- \,\frac{x}{1\,+\,x^2} \;= \;-\left[2\,+\,\frac{9x^2}{2(2\,-\,x^3)} \,+\,\frac{x}{1\,+\,x^2}\right]\)


Multiply by \(\displaystyle y:\)
. . . . \(\displaystyle \L y'\;=\;-y\left[2\,+\,\frac{9x^2}{2(2\,-\,x^3)} \,+\,\frac{x}{1\,+\,x^2}\right]\)


Replace \(\displaystyle y:\)
. . . . \(\displaystyle \L y' \;= \;-\frac{e^{-2x}(2\,-\,x^3)^{\frac{3}{2}}}{\sqrt{1\,+\,x^2}}\left[2\,+\,\frac{9x^2}{2(2\,-\,x^3)} \,+\,\frac{x}{1\,+\,x^2}\right]\)