Thread: equation of the tangent line to the curve

1. equation of the tangent line to the curve

determine the equation of the tangent line to the curve
f(x)=4x^3(x-2)^4(6x-5)^5 at point (1,4)

can u just help to get F'(x) then i know how to plug 1 in f'(X).

2. Originally Posted by Max10
determine the equation of the tangent line to the curve
f(x)=4x^3(x-2)^4(6x-5)^5 at point (1,4)

can u just help to get F'(x) then i know how to plug 1 in f'(X).

use the product rule (with the chain rule where necessary) and note that:

if $\displaystyle f, g, \text{ and } h$ are functions of $\displaystyle x$, then

$\displaystyle \frac d{dx} [fgh] = f'gh + fg'h + fgh'$

here you can let $\displaystyle f = 4x^3$, $\displaystyle g = (x - 2)^4$, and $\displaystyle h = (6x - 5)^5$

and note that you need the chain rule to find $\displaystyle g'$ and $\displaystyle h'$

3. so is it f'(x) 12x^2 g'(x)=4(x-2)^3(x) h'(x)=5(6x-5)^4(6)

4. Originally Posted by Max10
so is it f'(x) 12x^2 g'(x)=4(x-2)^3(x) h'(x)=5(6x-5)^4(6)

5. Originally Posted by Jhevon

how come (x-2)^4

g'(x) 4(x-2)^3(x)