Find an equation of the tangent line to the given curve at 'a':
y=((e^x)/4)-x ; a=0

Question

Find an equation of the tangent line to the given curve at 'a':
y=((e^x)/4)-x ; a=0

anonymous · Answer

take the derivative, replace $x$ by 0 to find the slope
what you wrote is not very clear however
is it 
$$f(x)=\frac{e^x}{4}-x$$?

anonymous · Answer

Yes, that is correct

anonymous · Answer

$f(0)=\frac{1}{4}$ so your point is $(0,\frac{1}{4})$ 
$$f'(x)=\frac{1}{4}e^x-1$$
$$f'(0)$$ is your slope, then use the point slope formula

anonymous · Answer

Thank you. Now, the next question is the same type of question, but the equation is $$y= e ^{x}$$ with a=ln3.  

I'm still trying to find the equation of the tangent line of the curve at 'a'. I know the answer, but I don't know how to get to it.

anonymous · Answer

dervivative of $e^x$ is $e^x$ 
your point is $(\ln(3), 3)$ since $e^{\ln(2)}=3$

anonymous · Answer

and similarly your slope is $3$