Find the line that is tangent to the curve y=x^3 at the point (1,1)
The deravitive of the function is f'(x)=-3X^2

Question

Find the line that is tangent to the curve y=x^3 at the point (1,1)
The deravitive of the function is f'(x)=-3X^2

anonymous · Answer

Y-1=-3(x^2-1)
y=-3x^2+4
This is what I am getting. 
The real answer is y=3x+2

anonymous · Answer

actually the derivative is $f'(x)=3x^2$  and so at  $x=1$ the slope is $f'(1)=3$

anonymous · Answer

don't forget the derivative is not itself a slope, but rather a formula for a slope
plug in the $x$ value, in this case 1, and find the slope.  the slope is a number, not a formula

anonymous · Answer

therefore you are asked to find the equation of a line with slope $3$ through the point $(1,1)$
the point-slope formula will complete this 

but again i stress, $f'(x)=3x^2$ is not itself the slope, it is the formula that gives it

anonymous · Answer

for example at $(2,8)$ the slope is $f'(2)=3\times 2^2=12$ etc

anonymous · Answer

I think I understand. Thanks for fixing both my errors.

anonymous · Answer

yw