use product rule to differentiate function

**JontyP** · March 2nd 2008, 05:34 AM

y= 7x^2 cos3x (with respect to x)

i got as far as -sin3x and 14x but got stuck here!

**janvdl** · March 2nd 2008, 05:43 AM

Originally Posted by JontyP

y= 7x^2 cos3x (with respect to x)

i got as far as -sin3x and 14x but got stuck here!

The product rule states that:

$(fg)(x) = f'(x)g(x) + f(x)g'(x)$

**janvdl** · March 2nd 2008, 05:46 AM

Where $f$ is the $7x^2$ part and $g$ is the $cos 3x$ part

**JontyP** · March 2nd 2008, 06:34 AM

i am using the other form of that
if y = u * v
so dy/dx = u (dv/dx) + v (du/dx)

and taking it from there. after finding the derivatives

y = 7 x^2 cos3x

i found the derivative of cos3x to be -3sin3x but cant get 7x^2 which is confusing

**janvdl** · March 2nd 2008, 06:39 AM

Originally Posted by JontyP

i am using the other form of that
if y = u * v
so dy/dx = u (dv/dx) + v (du/dx)

and taking it from there. after finding the derivatives

y = 7 x^2 cos3x

i found the derivative of cos3x to be -3sin3x but cant get 7x^2 which is confusing

Okay you can use that form. (It's the same thing anyway, and you should be able to recognise and use all the ways of writing it)

I'm taking your word for it that the derivative of cos 3x is -3sin 3x (Although it doesn't seem quite correct...)

Then:

$\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}$

$= (cos \ 3x) (14x) + (-3 sin \ 3x) (7x^2)$

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