Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Year 12s: what will be the first way you research your future options? >>

Need help with further differentiation please!

\frac{4x}{x^3}

-

\frac{3x^1/2}{x^3}

First time using latex, anyway please help.

(edited 11 years ago)

OP

\displaystyle \frac{4x}{\partial x} x^2y = 2xy

Original post by Tynos

\frac{4x}{x^3}

-

\frac{3x^1/2}{x^3}

First time using latex, anyway please help.

Write it in the form:

4x^p-3x^q

where you have to find p and q.

After doing so, differentiate.

Hint:

\frac{4}{x}=4x^{-1}

OP

I know lol, im stuck at the second part.

Hi, I don't know how to use the Math Editor so bear with me.
d/dx(x^p)=p*x^(p-1) provided p =/= 0
It's a fairly simple rule, the derivation is quite straightforward if you know binomial expansions.
If you use that with the hints above you can get an answer.

OP

Im just stuck on this part

\frac{3x^1/2}{x^3}

Original post by Tynos

Im just stuck on this part

\frac{3x^1/2}{x^3}

\frac{1}{2} - 3 = -\frac{5}{2}

so

\dfrac{3x^{\frac{1}{2}}}{x^3} = 3x^{-\frac{5}{2}}

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you

What are university entry requirements?

What are university entry requirements?

What university is really like

Where to look for financial support to help you study law

Where to look for financial support to help you study law

Applying to university: a simple guide

Applying to university: a simple guide