# Differentiating...

• Aug 5th 2010, 11:06 AM
bobsanchez
Differentiating...
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?
• Aug 5th 2010, 11:23 AM
Quote:

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...

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

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

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

is fairly compact
• Aug 5th 2010, 11:52 AM
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.
• Aug 5th 2010, 11:58 AM
skeeter
Quote:

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.
• Aug 5th 2010, 12:02 PM
bobsanchez
what happens with the 12?
• Aug 5th 2010, 12:10 PM
skeeter
Quote:

Originally Posted by bobsanchez
what happens with the 12?

what's the derivative of a constant?
• Aug 5th 2010, 12:17 PM
bobsanchez
Okay, so 17cos(x)cos(y) - 17sin(x)sin(y) = 0?
• Aug 5th 2010, 12:22 PM
Quote:

Originally Posted by bobsanchez
Okay, so 17cos(x)cos(y) - 17sin(x)sin(y) = 0?

You've lost the derivative!

$\displaystyle 17cosxsiny=12$

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

If differentiating wrt x

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

using the chain rule
• Aug 5th 2010, 12:29 PM
skeeter
$\displaystyle \displaystyle \frac{d}{dx}\left[17\cos{x}\sin{y} = 12\right]$

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

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

$\displaystyle \displaystyle \frac{dy}{dx} = \frac{\sin{x}\sin{y}}{\cos{x}\cos{y}} = \tan{x}\tan{y}$