# How do you find the slope of the tangent line to the graph of #y = (ln x) e^x# at the point where x = 2?

The slope of the tangent is

The gradient of the tangent to a function at any particular point is given by the derivative of the function at that point.

Differentiating

# \ \ \ \ \ dy/dx = (lnx)(d/dxe^x) + (d/dxlnx)(e^x) #

# :. dy/dx = (lnx)e^x + (1/x)(e^x) #

# :. dy/dx = ((lnx)+1/x)e^x#

When

# f'(2) = ((ln2)+1/2)e^2 = e^2ln2+1/2e^2#

If we wanted to find the equation of the tangent:

# \ \ f(2) =(ln2)e^2 = e^2ln2#

So the tangent we seek passes through

# y - e^2ln2 = (e^2ln2+1/2e^2 )(x-2) #

# :. y - e^2ln2 = (e^2ln2+1/2e^2 )x-2(e^2ln2+1/2e^2 ) #

# :. y - e^2ln2 = (e^2ln2+1/2e^2 )x-2e^2ln2-e^2 #

# :. y= (e^2ln2+1/2e^2 )x-e^2ln2-e^2 #

We can verify this is correct by the following graph:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the slope of the tangent line to the graph of y = (ln x) e^x at the point where x = 2, we can use the product rule and the chain rule of differentiation.

First, we differentiate the function y = (ln x) e^x using the product rule:

dy/dx = (d/dx)(ln x) * e^x + (ln x) * (d/dx)(e^x)

Next, we differentiate ln x and e^x using the chain rule:

(d/dx)(ln x) = 1/x (d/dx)(e^x) = e^x

Substituting these derivatives back into the expression for dy/dx, we have:

dy/dx = (1/x) * e^x + (ln x) * e^x

To find the slope of the tangent line at x = 2, we substitute x = 2 into the expression for dy/dx:

dy/dx = (1/2) * e^2 + (ln 2) * e^2

Therefore, the slope of the tangent line to the graph of y = (ln x) e^x at the point where x = 2 is (1/2) * e^2 + (ln 2) * e^2.

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- Find the equation of the line tangent to the parabola #y=x^2 -3x + 2# and perpendicular to the line #5x - 2y - 10=0# ?
- This is the second of three derivative problems. Can someone please help solve it?
- What is the equation of the line tangent to # f(x)=cscx # at # x=pi/2 #?
- What is the instantaneous rate of change of #f(x)=1/(x^2-x+3 )# at #x=0 #?
- Using the limit definition, how do you find the derivative of # f(x)=cosx #?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7