What is the equation of the tangent line of y=x^3-x^2e^(2x) at x=1?

1 Answer
Jun 15, 2017

y=32.5562x-38.9453 or if you need the exact answer

y=(3+4e^2)x-(2+5e^2)

Explanation:

Recall the point-slope formula:

y-y_1=m(x-x_1)

Where m=dy/dx

When x=1, we can plug that into the original equation for y

y=x^3-x^2e^(2x)

y=(1)^3-(1)^2e^(2*1)

y=1-e^2~~-6.3891

Finding the derivative of the equation for y, we get

dy/dx=d/dx(x^3)-d/dx(-x^2e^(2x))

dy/dx=3x^2-(-x^2 d/dx(e^(2x))+d/dx(-x^2)e^(2x))

dy/dx=3x^2+x^2(2e^(2x))+2xe^(2x)

When x=1, the slope of the tangent line is

dy/dx=3(1)^2+(1)^2(2e^(2(1)))+2(1)e^(2(1))

dy/dx=m=3+2e^(2)+2e^(2)=3+4e^2~~32.5562

Plugging slope, m=32.5562, the given x=1, and the derived y=-6.3891, all into the point slope formula, we get

y-y_1=m(x-x_1)

y+6.3891=32.5562(x-1)

y=32.5562x-32.5562-6.3891

y=32.5562x-38.9453

Or, if you need to be exact

y-1+e^2=(3+4e^2)(x-1)

y=3x-3+4xe^2-4e^2+1-e^2

y=(3+4e^2)x-(2+5e^2)