Equation of the tangent line...

[nikchic5]

If anyone could help me. I really need to understand how to do this..I have a Calc test on Monday! Thanks so much!

Find an equation of the tangent line to the curve at a given point.

y= (e^x) cos x, at the point (0,1)

again thanks so much for any help!

 

[pka]

If \(\displaystyle y = e^x \cos (x)\quad \Rightarrow \quad y' = e^x \cos (x) - e^x \sin (x)\), now the value of \(\displaystyle y'(0)\) is the slope of the tangent line.

Thus the answer is: \(\displaystyle y - 1 = y'(0)(x - 0)\)

 

[nikchic5]

Thanks

Thank you very much!