determining the equation of tangent line given ex^2x at x=1

May 2012
277
1
Toronto
the function is \(\displaystyle f(x)=xe^2x\)

find equation of tangent line at x=1

this is what I did

\(\displaystyle f'(x)=x*(e^2x)*2 + e^2x\)

\(\displaystyle f'(1)=3*e^2\)

so the equation of tangent line is

\(\displaystyle y=3x*e^2\)

is this correct?
 
Dec 2011
2,314
916
St. Augustine, FL.
First, a LaTeX tip: if your exponent contains more than 1 character, enclose it within braces, e.g., e^{2x}

You have correctly found \(\displaystyle f'(1)\), now for the tangent line, you want to use the point-slope formula:

\(\displaystyle y-f(1)=f'(1)(x-1)\)
 
May 2012
277
1
Toronto
test to see if I got the code right for TEX

the function is \(\displaystyle f(x)=xe^{2x}\)

find equation of tangent line at x=1

this is what I did

\(\displaystyle f'(x)=x*(e^{2x})*2 + e^{2x}\)

\(\displaystyle f'(1)=3*e^2\)

so the equation of tangent line is

\(\displaystyle y=3x*e^2\)

is this correct?