Question

i want to differentiate an equation...y=cX^(5/3)/(1+0.2x)^(2/3)...here c is the constant...can anyone help me?

Answer 1

I only studied the basics but im gonna try it

Answer 2

There is an edit in the question...i want to differentiate an equation...y=cX^(5/3)/(1+0.2x)^(2/3)-Q...here c &Q are constants...can anyone help me?

Answer 3

..?

Answer 4

no...
[y={cx^(5/3)/(1+0.2x)^(2/3)}-Q\]

Answer 5

Ok my mistake :)
\[\huge y=\frac{cx^{5/3}}{(1+0.2x)^{2/3}}-Q\]

Answer 6

So we need to apply the quotient rule.

Answer 7

\[\large y'=\frac{\color{orangered}{(\color{black}{cx^{5/3}})'}(1+0.2x)^{2/3}-cx^{5/3}\color{orangered}{\left(\color{black}{(1+0.2x)^{2/3}}\right)'}}{\left((1+0.2x)^{2/3}\right)^2}-0\]

Answer 8

Understand how I set it up? This one is a bit tricky.
This is the quotient rule setup.

The red terms are the ones we need to take a derivative of.

Answer 9

The -0 came from the Q, since it was constant.

Answer 10

This is the Quotient Rule for Derivatives, just in case you forget.
\[\huge \left(\frac{u}{v}\right)=\frac{u'v-uv'}{v^2}\]

Answer 11

Thanks a lot....