differentiation...gah please help

cliste09 · #1 $Old$ September 16th, 2009, 11:41 AM

hey this is an exam question i got stuck on,(im prob going to be posting a few of these just to let you know in advance lol) :

differentiate (x^2 + 9) (4x^3 + 5) with respect to x,

i know its where u have to use the dy over dx formula but i just have a hard time with this one and many others,so an easy explanation would be greatly appreciated as i am barely hanging on by a thread in ordinary level maths, the highest grade i have ever gotten is a C lol thanks!!

the correct answer is :20x^4 + 108x^2 + 10x

Matt Westwood · #2 $Old$ September 16th, 2009, 11:52 AM

Quote:

Originally Posted by cliste09 $View Post$

hey this is an exam question i got stuck on,(im prob going to be posting a few of these just to let you know in advance lol) :

differentiate (x^2 + 9) (4x^3 + 5) with respect to x,

i know its where u have to use the dy over dx formula but i just have a hard time with this one and many others,so an easy explanation would be greatly appreciated as i am barely hanging on by a thread in ordinary level maths, the highest grade i have ever gotten is a C lol thanks!!

the correct answer is :20x^4 + 108x^2 + 10x

Do you know the product rule? The diff of one thing multiplied by another thing is equal to the one thing multiplied by the diff of the other added to the diff of the one thing multplied by the other:

$\frac d {dx} f(x)g(x) = g (x) \frac d {dx} f(x) + f(x) \frac d {dx} g(x)$

That should get you started. Now all you need to do is work out $\frac d {dx} (x^2 + 9)$ and $\frac d {dx} (4x^3 + 5)$ .

If you're not sure how to do that then another hint: "the power law" which you ought to make sure you've learned.