Differentiating...

**bobsanchez** · August 5th 2010, 11:06 AM

Okay, so I'm finding the derivative of y = (5x - 1)^2 (3x + 7)^3

I've gotten it to y' = (5x - 1)^2 times 9(3x + 7)^2 + 10(5x-1) times (3x + 7)^3

My question is: do I combine all of these terms together or do you typically just leave it in this form?

**Archie Meade** · August 5th 2010, 11:23 AM

Originally Posted by bobsanchez

Okay, so I'm finding the derivative of y = (5x - 1)^2 (3x + 7)^3

I've gotten it to y' = (5x - 1)^2 times 9(3x + 7)^2 + 10(5x-1) times (3x + 7)^3

My question is: do I combine all of these terms together or do you typically just leave it in this form?

It may not be the most compact way to express the derivative,
but if you substitute a value of x into that, you will get the curve gradient at that x.

You can simplify further...

$10(5x-1)(3x+7)^3+9(3x+7)^2(5x-1)^2$

has $(3x+7)^2(5x-1)$ common

$(3x+7)^2(5x-1)\left(10[3x+7]+9[5x-1]\right)=(3x+7)^2(5x-1)(75x+61)$

is fairly compact

**bobsanchez** · August 5th 2010, 11:52 AM

Another similar "procedural" type question: is it necessary to put in terms of y in order to differentiate? For instance...

Differentiate 17cos(x)sin(y) = 12

Would I have to put in terms of y and, if so, how would I do that? If I can simply differentiate straightforwardly, I think I can do it.

**skeeter** · August 5th 2010, 11:58 AM

Originally Posted by bobsanchez

Another similar "procedural" type question: is it necessary to put in terms of y in order to differentiate? For instance...

Differentiate 17cos(x)sin(y) = 12

Would I have to put in terms of y and, if so, how would I do that? If I can simply differentiate straightforwardly, I think I can do it.

I recommend you use implicit differentiation.

**bobsanchez** · August 5th 2010, 12:02 PM

what happens with the 12?

**skeeter** · August 5th 2010, 12:10 PM

Originally Posted by bobsanchez

what happens with the 12?

what's the derivative of a constant?

**bobsanchez** · August 5th 2010, 12:17 PM

Okay, so 17cos(x)cos(y) - 17sin(x)sin(y) = 0?

**Archie Meade** · August 5th 2010, 12:22 PM

Originally Posted by bobsanchez

Okay, so 17cos(x)cos(y) - 17sin(x)sin(y) = 0?

You've lost the derivative!

$17cosxsiny=12$

$cosxsiny=\frac{12}{17}$

If differentiating wrt x

$siny(-sinx)+cosx\frac{dy}{dx}\frac{d}{dy}siny=0$

using the chain rule

**skeeter** · August 5th 2010, 12:29 PM

$\displaystyle \frac{d}{dx}\left[17\cos{x}\sin{y} = 12\right]<br />$

$\displaystyle \frac{d}{dx}\left[\cos{x}\sin{y} = \frac{12}{17}\right]<br />$

$\displaystyle<br /> \cos{x}\cos{y} \cdot \frac{dy}{dx} + \sin{y}(-\sin{x}) = 0$

$\displaystyle<br /> \frac{dy}{dx} = \frac{\sin{x}\sin{y}}{\cos{x}\cos{y}} = \tan{x}\tan{y}$

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