# Math Help - [SOLVED] Gradient of a function

1. ## [SOLVED] Gradient of a function

Find the gradient of the function 3sqrt theta^3 / 2 Sin 2 theta giving your answer as a product of factors.

I have looked at this equation and as far as can tell I need to use Chain rule, product rule and quotient rule.

Is this right? and if so in what order should i use them?

2. Originally Posted by ally79
Find the gradient of the function 3sqrt theta^3 / 2 Sin 2 theta giving your answer as a product of factors.

I have looked at this equation and as far as can tell I need to use Chain rule, product rule and quotient rule.

Is this right? and if so in what order should i use them?

Do you mean $f(\vartheta)=\frac{3\sqrt{\vartheta^3}}{2\sin(2\va rtheta)}$ or $f(\vartheta)=\frac{3}{2}\sqrt{\vartheta^3}\sin(2\v artheta)$??

--Chris

3. The first one

4. Originally Posted by ally79
Find the gradient of the function 3sqrt theta^3 / 2 Sin 2 theta giving your answer as a product of factors.

I have looked at this equation and as far as can tell I need to use Chain rule, product rule and quotient rule.

Is this right? and if so in what order should i use them?
$f(\vartheta)=\frac{3\sqrt{\vartheta^3}}{2\sin(2\va rtheta)}$

You will need to apply the chain rule and quotient rule here...

$f'(\vartheta)=\frac{\displaystyle \left[2\sin(2\vartheta)\cdot\frac{3}{2}(\vartheta^3)^{-\frac{1}{2}}\cdot3\vartheta^2\right]-\left[3\sqrt{\vartheta^3}\cdot 4\cos(2\vartheta)\right]}{4\sin^2(2\vartheta)}$ $=\color{red}\boxed{\frac{9\vartheta^2\sin(2\varthe ta)-12\vartheta^3\cos(2\vartheta)}{4\sqrt{\vartheta^3} \sin^2(2\vartheta)}}$

I hope this makes sense!

--Chris

5. Hi,

Another method that can maybe simplify things :

$\sqrt{\theta^3}=(\theta^3)^{\tfrac 12}=\theta^{\tfrac 32}$. The derivative of $x^n$ is $nx^{n-1}$, so the derivative of $\theta^{\tfrac 32}$ is $\tfrac 32 \theta^{\tfrac 12}=\tfrac 32 \sqrt{\theta}$
And the derivative of $a \cdot P(x)$, where a is a constant is $a \cdot P'(x)$. So we can pull out the $\tfrac 32$ factor.

$f(\theta)=\frac 32 \cdot \frac{\theta^{\tfrac 32}}{\sin(2 \theta)}$

Quotient rule:

$f'(\theta)=\frac 32 \cdot \frac{\tfrac 32 \sqrt{\theta} \cdot \sin(2 \theta)-\sqrt{\theta^3} \cdot 2 \cos(2 \theta)}{\sin^2(2\theta)}$

$f'(\theta)=\frac 34 \cdot \frac{3 \sqrt{\theta} \sin(2 \theta)-4 \sqrt{\theta^3} \cos(2 \theta)}{\sin^2(2\theta)}$

6. Hi,
I followed the example by chris L t521 and managed to work it out. However i'm a little unclear as to how he simplifies it. If someone could help me with that it would be great

Cheers

Ally

7. Originally Posted by Chris L T521
$f(\vartheta)=\frac{3\sqrt{\vartheta^3}}{2\sin(2\va rtheta)}$

You will need to apply the chain rule and quotient rule here...

$f'(\vartheta)=\frac{\displaystyle \left[2\sin(2\vartheta)\cdot\frac{3}{2}(\vartheta^3)^{-\frac{1}{2}}\cdot3\vartheta^2\right]-\left[3\sqrt{\vartheta^3}\cdot 4\cos(2\vartheta)\right]}{4\sin^2(2\vartheta)}$ $=\color{red}\boxed{\frac{9\vartheta^2\sin(2\varthe ta)-12\vartheta^3\cos(2\vartheta)}{4\sqrt{\vartheta^3} \sin^2(2\vartheta)}}$

I hope this makes sense!

--Chris
Originally Posted by ally79
Hi,
I followed the example by chris L t521 and managed to work it out. However i'm a little unclear as to how he simplifies it. If someone could help me with that it would be great

Cheers

Ally
Sorry...I left it for you to try to simplify. Now that you're stuck, I'll now include the extra steps I left out.

$\frac{\displaystyle \left[2\sin(2\vartheta)\cdot\frac{3}{2}(\vartheta^3)^{-\frac{1}{2}}\cdot3\vartheta^2\right]-\left[3\sqrt{\vartheta^3}\cdot 4\cos(2\vartheta)\right]}{4\sin^2(2\vartheta)}$

Let's focus on the first grouped part in the numerator:

$2\sin(2\vartheta)\cdot\frac{3}{2}(\vartheta^3)^{-\frac{1}{2}}\cdot3\vartheta^2)$

We see a something cancels out, and we can combine a couple terms:

$\not 2\sin(2\vartheta)\cdot\frac{{\color{red}3}}{\not 2}(\vartheta^{-\frac{3}{2}}\cdot{\color{red}3}\vartheta^2)=9\sin( 2\vartheta)(\vartheta^{-\frac{3}{2}}\cdot\vartheta^2)$

Now when subtracted from the second term in the numerator, we get

$9\sin(2\vartheta)({\color{red}\vartheta^{-\frac{3}{2}}}\cdot\vartheta^2)-12{\color{red}\sqrt{\vartheta^3}}\cos(2\vartheta)$

The common factor to both terms is $\vartheta^{-\frac{3}{2}}$, but this is what we get when we pull it out:

$\vartheta^{-\frac{3}{2}}\left(9\sin(2\vartheta)\vartheta^2-12\vartheta^3\cos(2\vartheta)\right)$

Recall that the denominator of the derivative was $4\sin^2(2\vartheta)$

So we now see that $f'(\vartheta)=\frac{\displaystyle\vartheta^{-\frac{3}{2}}\left(9\vartheta^2\sin(2\vartheta)-12\vartheta^3\cos(2\vartheta)\right)}{4\sin^2(2\va rtheta)}$

This is the same as saying $\color{red}\boxed{f'(\vartheta)=\frac{\displaystyl e9\vartheta^2\sin(2\vartheta)-12\vartheta^3\cos(2\vartheta)}{4\sqrt{\vartheta^3} \sin^2(2\vartheta)}=\frac{\displaystyle9\sqrt{\var theta}\sin(2\vartheta)-12\sqrt{\vartheta^3}\cos(2\vartheta)}{4\sin^2(2\va rtheta)}}$

I think that the last one is even better...

Does this clarify things?

--Chris

8. Pleas excuse my ignorance, I followed everything you did and understood it, except for one part.

When you take the common factor of theta^-3/2 from both sides

You have sqrt(theta^3) I was told that that was the same as theta^3/2 not negative 3/2

Again i apologise if i'm being ignorant

9. Originally Posted by ally79
Pleas excuse my ignorance, I followed everything you did and understood it, except for one part.

When you take the common factor of theta^-3/2 from both sides

You have sqrt(theta^3) I was told that that was the same as theta^3/2 not negative 3/2

Again i apologise if i'm being ignorant
Your correct, but keep in mind that $\vartheta^{\frac{3}{2}}=\frac{\displaystyle\varthe ta^3}{\displaystyle\vartheta^{\frac{3}{2}}}=\varth eta^3\cdot\vartheta^{-\frac{3}{2}}$