How do I get from the problem to the answer using the chain rule?
It would be much appreciated if someone indicated the steps for me, thank you!
$\displaystyle y'=2x(x^2+2)^\frac{1}{3}+(x^2+1)\cdot\frac{1}{3}(x ^2+2)^\frac{-2}{3}(2x)$
rewriting the second term with a positive exponent gives
$\displaystyle y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\cdot\frac{1}{ 3(x^2+2)^\frac{2}{3}}$
multiplying the numerator and denominator of the second term
[tex]MATH]
simplifying
$\displaystyle y'=2x(x^2+2)^\frac{1}{3}+(2x)(x^2+1)\frac{(x^2+2)^ \frac{1}{3}}{3(x^2+2)}$
factoring out the GCF gives
$\displaystyle y'=2x(x^2+2)^\frac{1}{3}\left[1+\frac{x^2+1}{3(x^2+2)}\right]$
Thank you very much, everyone. Now I understand how to do that problem.
I also have trouble determining the derivatives of these and I'm not sure if I'm doing them correctly because the book don't have their answers.
$\displaystyle 1. y = x sin x^{1/2}$
$\displaystyle y' = x cos (x^{1/2}) d/dx (x^{1/2}) + (sin x^{1/2}) d/dx (x^{1/2})$
$\displaystyle y' = x cos \sqrt{x} + sin \sqrt{x} / {2x \sqrt{x}}$
2. $\displaystyle y = x/(7-3x^{1/2})$
I only got up to $\displaystyle (7-3x)^{1/2} (-3) - x[(1/2)(7-3x^{1/2})] (-3)$
1. You need the product rule:
$\displaystyle u = x \Rightarrow \frac{du}{dx} = 1$
$\displaystyle v = \sin x^{1/2} \Rightarrow \frac{dv}{dx} = \cos x^{1/2} \times \frac{1}{2} x^{-1/2} = \cos \sqrt{x} \times \frac{1}{2\sqrt{x}}$, where the chain rule has also been used.
Then you substitute the above results into $\displaystyle \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$.
Some simplification of the resulting answer will be possible .....
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2. You need the quotient rule:
$\displaystyle u = x \Rightarrow \frac{du}{dx} = 1$
$\displaystyle v = 7 - 3 x^{1/2} \Rightarrow \frac{dv}{dx} = -\frac{3}{2} x^{-1/2} = -\frac{3}{2\sqrt{x}}$.
Then you substitute the above results into $\displaystyle \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$.
Some simplification (requiring a little bit of algebra) of the resulting answer will be desirable .....