Find the gradient of the tangent to the curve y=(3x-1)^3 (2x-1)^2 at the point where x=2

Question

anonymous · Answer

take the derivative, probably using the product and chain rule
then replace $x$ by $2$

anonymous · Answer

you got the derivative, or you need help with it?  
no need so simplify using any algebra, because your job will then be to evaluate at $x=2$ so you can do that no matter what form the derivative is in

anonymous · Answer

$$y=(3x-1)^3 (2x-1)^2 $$
$$y'=3\times 3(3x-1)^2(2x-1)+2\times 2(x-1)(3x-1)^3$$  by the product and chain rule
just leave it, don't simplify
replace $x=2$ and do the arithmetic

anonymous · Answer

3(3 x-1)^2 2(2x-1)=450

anonymous · Answer

well maybe simplify to 
$$y'=9(3x-1)^2(2x-1)+4(x-1)(3x-1)^3$$

anonymous · Answer

hold on
that is not the derivative
you have a product
use $(fg)'=f'g+g'f$ with 
$$f(x)=(3x-1)^3, f'(x)=9(x-1)^2, g(x)=(2x-1)^2, g'(x)=4(x-1)$$

anonymous · Answer

I am still not getting the answer

anonymous · Answer

Its 3525 but i am getting 1175

anonymous · Answer

$$(2 x-1)^2 (3 x-1)^3=108 x^5-216 x^4+171 x^3-67 x^2+13 x-1 $$The derivative of the RHS above is$$540 x^4-864 x^3+513 x^2-134 x+13 $$and when evaluated at x=2 is equal to 3525.