Can someone please help me find the derivative of this problem? I will fan and medal.

- anonymous

Can someone please help me find the derivative of this problem? I will fan and medal.

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- schrodinger

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- Jhannybean

Whats the function?

- anonymous

I'm typing it

- anonymous

\[\frac{ 2 }{ 3 }\sin^\frac{ 3 }{ 2 }x-\frac{ 2 }{ 7 }\sin^\frac{ 7 }{ 2 }x\]

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## More answers

- Jhannybean

\[\frac{2}{3}\left(\sin^{2/3}x\right)' \iff \frac{2}{3}\left((\sin(x))^{2/3}\right)'\]

- wmj259

If you look carefully you see that the two functions are connected with the subtract sign. So you can break that function into 2 separate derivatives.

- wmj259

\[\frac{ d }{ dx } (\frac{ 2 }{ 3 }\sin ^{2/3}(x)) - \frac{ d }{ dx }((\frac{ 2 }{ 7 }\sin ^{7/2}x))\]

- Jhannybean

Use the power rule, then the chain rule.

- wmj259

|dw:1444350991307:dw|

- Jhannybean

lmao -_-

- anonymous

Wait. How did you guys turn the exponent 3/2 into 2/3?

- Jhannybean

Typo.

- wmj259

yes that was a typo on my part, I apologize.

- anonymous

Y'all are fine. I was just making sure.

- Jhannybean

\[\large \frac{d}{dx}\left[\frac{2}{3}(\sin(x))^{3/2}\right] = \frac{2}{3} \cdot \frac{3}{2} \cdot (\sin(x))^{(3/2) - 1} \cdot \frac{d}{dx} (\sin(x))\]

- wmj259

What I would do in this situation would be to first split the problem into 2 different differentiation's because of that subtract sign. You can pull out the (2/3) and the (2/7) because they are constants
And then I would do the Power Rule on the (3/2) and the (7/2). Separately.
And then I would do the chain rule on the 2 sin(x) functions.

- wmj259

|dw:1444351292444:dw|

- Jhannybean

\[\large \frac{d}{dx}\left[\frac{2}{7}(\sin(x))^{7/2}\right] = \frac{2}{7} \cdot \frac{7}{2} \cdot (\sin(x))^{(7/2)-1} \cdot \frac{d}{dx}(\sin(x))\]

- wmj259

|dw:1444351320870:dw|

- wmj259

|dw:1444351344236:dw|

- anonymous

\[\sin^\frac{ 1 }{ 2 }x*cosx-\sin^\frac{ 5 }{ 2 }x*cosx\]

- anonymous

That's what I got

- Jhannybean

For which part?

- anonymous

The entire problem

- Jhannybean

Alright let's check

- anonymous

I just did the power rule then chain rule as recommended.

- wmj259

|dw:1444351512241:dw|

- wmj259

Is the finished derivative for the left half.

- wmj259

the simplified left side is
|dw:1444351605530:dw|

- wmj259

Now the right side would give us:
|dw:1444351641222:dw|

- wmj259

The simplified right side would be |dw:1444351722466:dw|

- wmj259

So if you combine the left and the right side you get:
|dw:1444351765027:dw|

- wmj259

@Catseyeglint911, Which is what you got.

- anonymous

@wmj259
Can you simplify the right side any further?

- wmj259

you can say this:
|dw:1444351917329:dw|

- wmj259

|dw:1444351988066:dw|

- Jhannybean

\[\begin{align}\large \frac{d}{dx}\left[\frac{2}{3}(\sin(x))^{3/2}\right] &= \frac{2}{3} \cdot \frac{3}{2} \cdot (\sin(x))^{(3/2) - 1} \cdot \frac{d}{dx} (\sin(x)) \\&= \sin^{1/2}(x)\cos(x) \ \end{align}\]\[\begin{align} \large \frac{d}{dx}\left[\frac{2}{7}(\sin(x))^{7/2}\right] &= \frac{2}{7} \cdot \frac{7}{2} \cdot (\sin(x))^{(7/2)-1} \cdot \frac{d}{dx}(\sin(x)) \\&=\sin^{5/2}(x)\cos(x)\end{align}\]\[\begin{align}\frac{d}{dx}\left[ \frac{ 2 }{ 3 }\sin^\frac{ 3 }{ 2 }x-\frac{ 2 }{ 7 }\sin^\frac{ 7 }{ 2 }x\right] &= \sin^{1/2}(x)\cos(x) -\sin^{5/2}(x)\cos(x) \\&= \sin^{1/2}(x)\cos(x) \left[ 1-\sin^{5}(x) \right] \end{align} \]

- Jhannybean

\[\large \color{red}{\boxed{\sin^{1/2}(x)\cos(x)\left[1-\sin^{5}(x)\right]}}\]

- Jhannybean

@wmj259 you did something funky here|dw:1444352122644:dw|

- wmj259

Yes, The right most part should have been the derivative of sin(x) not just (x).

- wmj259

It goes much longer then that because their is basically a Power rule then 2 chain rules. but the 2nd chain rule is always ignore because its just the derivative of x's in some cases. But with practice comes great speed.

- anonymous

Thank you both! I wish I could give out multiple medals. I'll just become a fan of both of you. :)

- Jhannybean

Woohoo!

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