Differentiation with the chain rule

[the tree]

I was doing an exercise diffentiating things like (2+3x)⁴, which was fairly easy, being given the rule: if y=[f(x)]ⁿ then (dy/dx)=n[f(x)]^n-1f'(x)

 

But then, to my horror, another number turned up that I didn't know what to do with, there was a question like so: 5(2+9x)³ and I just don't know what to do here!

 

Could anyone give me a nudge in the right direction?

[timo]

What is your problem? The factor 5? Differentiation is a linear operation, therefore d/dx (5 f(x)) = 5 (d/dx f(x)).

 

EDIT: Alternatively, use the product rule: d/dx (5 f(x)) = ( d/dx 5) f(x) + 5 (d/dx f(x)) = 0 + 5 (d/dx f(x))

[the tree]

Thanks, that's exactly what I needed to know. I thought that might be what I was meant to do, but I had no way of checking.

[Ragib]

think of it intuitively, if theres a factor of 5, then on a graph, every y value is just 5 times as much as the original. Infintesimal changes wud also be 5 times as much, making the derivative 5 times as much as the function without the factor 5. I know, my typing is bad...We'll i find that when i can't do something one way, i do it the longer way..i would have just expanded the terms then used the power rule, then maybe you would have noticed it was just 5 times as much