Find the equation of the tangent line to the curve ln(xy)=2x at the point (1,e^2).

Question

Tangent line y=______________

Answer 1

ln(xy) = 2x

xy = e^2x

y= e^2x/x =(e^2x)(1/x) = (e^2x)(x^-1) use the product rule. the derivative of y is the 1st term times the derivative of the 2nd term plus the 2nd term times the derivative of the 1st term.

y' = -e^2x/x^2 + (1/x)(e^2x)(2)

at (1,e^2) y' = -e^2 +2e^2 = e^2

the tangent line has slope = e^2

y-e^2 = (e^2)(x-1)

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

y= (e^2)x is the tangent line. It goes through the origin with slope=about 7.4

y = about 7.4x

Answer 2

1. What is the general equation of a tangent line?

Recall the point slope form of the linear equation:

y-y₀=m(x-x₀), where m is the slope

We are given (x₀,y₀)= (1,e²). The key is to find the slope.

y-e²=m(x-1)

1. Express function ln(xy)=2x in terms of f(x) or y= ... x...

Let's move x to one side and y to another.

e ^ln(xy)=e^2x

xy=e^2x

y= (e^2x)/x

1. Find the slope of the function y= (e^2x)/x

The slope of the equation is the first derivative of the function y= (e^2x)/x

Does quotient rule ring a bell?

d(Hi)/d(H₀)=(H₀dHi-Hi*dH₀)/(H₀*H₀), where Hi= e^2x and H₀=x

*** notice the d(Hi) means to take the derivative of Hi, where d(H₀) is derivative of H₀***

Now do the pleasure of substitution:

d(e^2x ) /d(x) = [xd(e^2x)- e^2xd(x)]/(x²)

And the pleasure of taking derivatives:

(2xe^2x-e^2x)/(x²)= e^2x(2x-1)/x²

d(Hi)/d(H₀)=e^2x(2x-1)/x².... whola, here is the slope m!

1. Plug the slope back to the equation of tangent line

y-e²=m(x-1)

y-e²=e^2x(2x-1)/x² * (x-1)