Use the product rule to find the derivative and the find the equation of the tangent line
for the function f(x)=(x^2+x-3)(3x^3-8x+1 at the point x=-1

Question

Use the product rule to find the derivative and the find the equation of the tangent line
for the function f(x)=(x^2+x-3)(3x^3-8x+1 at the point x=-1

anonymous · Answer

The product rule is
d(f(x)*g(y))=f(x)*g'(y)+g(y)*f'(x)

Can you do the rest?

anonymous · Answer

do I just plug in the numbers

anonymous · Answer

Plugging x=-1 into the derivative will give you the corresponding y coordinate. You need to find enough information to be able to find an equation for the tangent line.

anonymous · Answer

so it would be $$(x^2+x-3)(3x^3-8x+1)=(x^2+x-3)(3x^3-8x+1)+(3x^3-x+1)(x^2+x-3) Right ?$$

anonymous · Answer

I'm not quite sure. Can you resend just the right hand side?

anonymous · Answer

$$(x^2+x-3)(3x^3-8x+1)=$$

anonymous · Answer

Yeah, so what's the derivative of that?

anonymous · Answer

I'm not sure what to do

anonymous · Answer

Let's say that $$f =x^2+x-3$$
and
$$g=3x^3-8x+1$$
The product rule says the derivative of $$f \times g=f \times g \prime + g \times f \prime$$
Let me know what you get