Question

Find an equation of the tangent line to the given curve at the specified point. y = ex x , (1, e)

Answer 1

Is y = \(\large e^xx\) ?

Answer 2

is that at the point (1,e)??

Answer 3

It is not cleat what y is and so I am asking if the function is e raised to x times x as shown in my first reply.

Answer 4

yes
it is

Answer 5

my bad

Answer 6

To find the equation of the tangent we need the slope of the tangent.
Find the derivative dy/dx first. What do you get?

Answer 7

e^x is the slope

Answer 8

Could you post a screenshot of the question?

Answer 9

and this question too please

Answer 10

what is the derivative of \(\frac{e^x}{x}\) ?

Answer 11

its (1/x - 1/x^2)e^x

Answer 12

The derivative can be simplified to: \(\large y' = \frac{e^x(x-1)}{x^2} \)
Evaluate the derivative at x = 1
y' = 0.
Therefore, the slope of the tangent to the curve at (1,e) is zero. That means the tangent is a horizontal line parallel to the x-axis since the slope is zero. Horizontal lines have the equation y = constant. And since this tangent passes through the point (1,e) where the y-value is 'e', the equation of the tangent is y = e.

Answer 13

that worked, thanks

Answer 14

For the second question, I will do one and you can try the rest.
a) h(x) = 5f(x) - 4g(x)
h'(2) = ?
h(x) = 5f(x) - 4g(x)
h'(x) = 5f'(x) - 4g'(x)
h'(2) = 5f'(2) - 4g'(2) = 5*(-3) - 4*1 = -15 - 4 = -19