1
$\begingroup$

This is a very simple problem, but I am stuck on one step:

Differentiate $(\frac{a+x}{a-x})^{\frac{3}{2}}$

Now, this is what I have done:

$$ (\frac{a+x}{a-x})^{\frac{3}{2}} \\ \implies \frac{\delta}{\delta y}\frac{f}{g} \\ \implies gf' = (a-x)^{\frac{3}{2}} \times \frac{3}{2} (a+x)^{\frac{1}{2}} \times 2 \\ \implies fg' \implies (a+x)^{\frac{3}{2}} \times \frac{3}{2} (a-x)^{\frac{1}{2}} \times 0 = 0 \\ \implies \frac{(a-x)^{\frac{3}{2}} \times 3 (a+x)^{\frac{1}{2}}}{(a-x)^3}\\ \implies \frac{(a-x)^{\frac{3}{2}} - 3\sqrt{a+x}}{(a-x)^3} $$

But the answer is: $$ \frac{3\times a (a+x)^{\frac{1}{3}}}{(a-x)^{\frac{5}{2}}} $$

WolframAlpha shows:

$$ \frac{3a \sqrt{\frac{a+x}{a-x}}}{(a-x)^2} $$

Another Answer (Somehow I got this):

$$ \frac{3 \sqrt{\frac{a+x}{a-x}}}{2(a-x)} $$

==================

EDIT 1: What about:

$$ y = (\frac{a+x}{a-x})^{\frac{3}{2}} \\ y = u^{\frac{3}{2}} \hspace{0.5cm} ; \hspace{0.5cm} u = \frac{a+x}{a-x}\\ \implies \frac{3}{2}u^{\frac{1}{2}} \hspace{0.5cm} ; \hspace{0.5cm} \frac{(0+1)\times (a-x) - [ -1 (a+x) ]}{(a-x)^2} \\ \implies \frac{2a}{(a-x)^2} \\ \implies \frac{3}{2}\sqrt{\frac{2a}{(a-x)^2}} = \frac{3}{2} \times \frac{\sqrt{2a}}{a-x} $$

$\endgroup$
7
  • $\begingroup$ Where did the $0$ come from? Derivatives of this form just require careful bookkeeping $\endgroup$
    – Chinny84
    Nov 5, 2015 at 17:28
  • $\begingroup$ You seem to have errors in your computations. The "$\times 2$" and "$\times 0$" are wrong, and the second one is missing a "$\times(-1)$". $\endgroup$
    – MPW
    Nov 5, 2015 at 17:30
  • $\begingroup$ Differentiating $(a-x) = 1\times a^{1-1} - 1\times x^{1-1} = 0$ that's how. And, similarly: $ (a+x) = 1.(a^{1-1}) + 1.(x^{1-1}) $ $\endgroup$
    – weirdpanda
    Nov 5, 2015 at 17:35
  • $\begingroup$ @Chinny84 Now I got it. We're treating $a$ as a constant, and since constants differentiate to 0... $\endgroup$
    – weirdpanda
    Nov 5, 2015 at 17:43
  • $\begingroup$ I understand that concept but you should have two terms for $fg'$ one for the derivative of $a$ and one for the non-constant $x$. Maybe I just misread what you did. $\endgroup$
    – Chinny84
    Nov 5, 2015 at 17:45

2 Answers 2

Reset to default
1
$\begingroup$

i think the right answer is this here $$\frac{3}{2} \sqrt{\frac{a+x}{a-x}} \left(\frac{a+x}{(a-x)^2}+\frac{1}{a-x}\right)$$

$\endgroup$
2
  • $\begingroup$ I have the first part of the answer; chain rule, then quotient rule, but the second part...? $\endgroup$
    – weirdpanda
    Nov 5, 2015 at 17:46
  • $\begingroup$ bring the second part together and you will see it $\endgroup$
    – Dr. Sonnhard Graubner
    Nov 5, 2015 at 17:47
1
$\begingroup$

So, there was something flawed in my calculation:

$$ y = f(g(x)) \\ y' = \frac{dy}{du} \times \frac{du}{dx} \\ $$

Let's see it again: $$ y = (\frac{a+x}{a-x})^{\frac{3}{2}} \\ y = u^{\frac{3}{2}} \hspace{0.5cm} ; \hspace{0.5cm} u = \frac{a+x}{a-x} \\ \frac{dy}{du} = \frac{3}{2} \times u^{\frac{1}{2}} \implies \frac{3}{2} \times (\frac{a+x}{a-x})^{\frac{1}{2}} \\ \frac{du}{dx} = \frac{2a}{(a-x)^2} \\ y' = \frac{dy}{du} \times \frac{du}{dx} \\ ...\\ \frac{3a(a+x)^{\frac{1}{2}}}{(a-x)^{\frac{5}{2}}} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .