# Differentiation; trouble finding $\frac{dy}{dx}$ in terms of $y$

I'm self-studying through some exercises on differentiation, and have found a section which I've gotten almost every question wrong. Can anyone help me find out where I'm going wrong?

An example of one of the questions and my attempt to answer it is

Q: Find $\frac{dy}{dx}$ in terms of $y$ when $x = 4y^5-8\sqrt{y}$

My answer: $$x = 4y^5-8\sqrt{y} = 4y^5 - 8y^\frac{1}{2}$$ $$\frac{dx}{dy} = 20y^4 - 4y^{-\frac{1}{2}}$$ $$= 20y^4-\frac{4}{\sqrt{y}}$$ since: $$\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} \\ \therefore \frac{dy}{dx} = \frac{1}{20y^4}-\frac {y^{-\frac{1}{2}}}{4} \\ = \frac {4 - 20(y^{-\frac{1}{2}})}{80y^4} \\ = \frac {1 - 5(y^{-\frac{1}{2}})}{20y^4}$$

But the listed answer is: $$\frac{y^\frac{1}{2}}{4(5y^\frac{9}{2}-1)}$$

I suspect that due to the fact I've gotten only 2 out of the 6 questions right I'm doing something fundamentally wrong! Can anyone help me figure out either where I'm going wrong, or what I'm missing?

The idea is right, but you have not taken the reciprocal correctly. In general $\frac{1}{a+b} \neq \frac{1}{a} + \frac{1}{b}$. For example, $\frac{1}{2+2} \neq \frac{1}{2} + \frac{1}{2} = 1$. The answer is just $$\frac{1}{20y^4-\frac{4}{\sqrt{y}}},$$ although the listed answer is simplified.

• Thank you for pointing this out. I can get to the correct answer from here. Dec 13, 2012 at 13:35

Since you are self-studying, perhaps you would like to attempt the question again, using Jair's advice.

If you get stuck again you can compare with the solution below.

\begin{eqnarray*} \frac{dy}{dx} & = & \left(\frac{dx}{dy}\right)^{-1} \\ & = & \frac{1}{20y^4 - 4y^{-1/2}} \\ \\ & = & \frac{y^{1/2}}{y^{1/2}} \cdot \frac{1}{20y^4 - 4y^{-1/2}} \\ \\ & = & \frac{y^{1/2}}{20y^{9/2} - 4} \\ \\ & = & \frac{y^{1/2}}{4(5y^{9/2} - 1)} \end{eqnarray*}

• Thanks, for taking the time to write a solution. I managed to get a solution (matching yours) from Jair's advice. Dec 13, 2012 at 13:39

# Hint:

You can differentiate both sides of $$x = 4y^5-8\sqrt{y}$$ by $$x$$, meaning that function $$y=y(x)$$ depends on $$x,$$ then calculate $$\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}.$$