Find an equation of the tangent line to y=f(x) at x=1
f(x)=3e^(x^2)
please help1!!

Question

anonymous · Answer

Have you found the derivative of the curve yet?

solomonzelman · Answer

1) Derive the equation.
2) Plug in the 1 for x, (since it is at x=1)
3) then the point is (1,f(1)).

solomonzelman · Answer

Any questions ?

solomonzelman · Answer

I mean plug in 1 for x, into the derivative.

solomonzelman · Answer

Step 1:
$\large\color{blue}{  y=3e^{x^2}  }$
$\large\color{blue}{  y'=3e^{x^2}\times2x=6xe^{x^2}  }$

solomonzelman · Answer

Step 2:
$\large\color{blue}{  y'=3e^{x^2}\times2x=6xe^{x^2}  }$
$\large\color{blue}{  m_{tan}=6(1)e^{1^2}=6e^1=6e  }$

solomonzelman · Answer

So you know the slope is 6e.

solomonzelman · Answer

Find f(1), which is the y-coordinate of the point.

solomonzelman · Answer

$\large\color{blue}{ f(x)=3e^{x^2}}$
$\large\color{blue}{ f(1)=3e^{1^2}=3e}$

So the point is (1,3e)

solomonzelman · Answer

$\large\color{blue}{ y-y_1=m(x-x_1)}$

$\large\color{blue}{ y-3e=6e(x-1)}$

$\large\color{blue}{ y-3e=6e~x~-~6e}$

$\large\color{blue}{ y=6e~x~-~3e}$

solomonzelman · Answer

This is the answer. (   $\large\color{blue}{ y=6e~x~-~3e}$   )

anonymous · Answer

Oh my, thank you sosososososo mcuh!!!!!!!!! :D