Product & Quotient Rules #6

Question

Find an equation of the line that is tangent to the curve  at the point 

Tangent Line Equation: 

Answer 1

Line equation: y-y₁ = m(x-x₁)

m = dy/dx = e^xsin(x) + e^xcos(x)

At (x,y) = (0,0)

m = e⁰sin(0) + e⁰cos(0) = 1

Tangent line equation: y - 0 = 1(x-0) --> y = x

Answer 2

dy/dx = e^x(cos x + sin x)

at x=0, the derivative is 1

and the tangent line is 1 = y/x, i.e. y=x

Answer 3

y' = e^xsinx + e^xcosx; slope of tangent line m = y'(0, 0) = e⁰sin0 + e⁰cos0 = 0 + 1 = 1.

So equation of tangent line y = 1(x - 0); y = x