What is the equation of the tangent line of #f(x) =e^x/x-e^(2x-3)/x^3# at #x=2#?

Answer 1

Use the quotient rule to differentiate both functions and then use the difference rule.

Let #g(x) = e^x/x#
#g'(x) = (e^x xx x - e^x xx 1)/x^2#
#g'(x) = (xe^x - e^x)/x^2#
#g'(x) = (e^x(x - 1))/x^2#
#:.# Let #h(x) = e^(2x - 3)/x^3#

We must differentiate the numerator using the chain rule--

Let #y = e^u#
#u = 2x - 3#
#y' = e^u#
#u' = 2#
#:.y' = e^(2x - 3) xx 2#

As for the denominator:

#y = x^3#
#y' = 3x^2#
Hence, #h'(x) = (2e^(2x - 3) xx x^3 -e^(2x - 3) xx 3x^2)/(x^3)^2#
#h'(x) = (2x^3e^(2x - 3) - 3x^2e^(2x -3))/x^6#
#h'(x) = (e^(2x - 3)(2x^3 - 3x^2))/x^6#
#:. f'(x) = (e^x(x - 1))/x^2 - (e^(2x - 3)(2x^3 - 3x^2))/x^6#
The slope of the tangent is given by evaluating #f'(a)#, a being the given point, #x = a#.

So, we have

#f'(2) = (e^2(2 - 1))/2^2 - (e^(2(2) - 3)(2(2)^3 - 3(2)^2))/2^6#
#f'(2) = e^2/4 - (4e)/64#
#f'(2) = e^2/4 - e/16#
#f'(2) = (4e^2 - e)/16#
The slope of the tangent is #(4e^2 - e)/16#.

We need to find the y value that the function and the tangent passes through.

#f(2) = e^2/2 - e^1/8#
#f(2) = (4e^2 - e)/8#

Now that we know a point on the function and on the tangent and the slope of the tangent, we can find the equation of the tangent.

#y - y_1 = m(x - x_1)#
#y - ((4e^2 - e)/8) = (4e^2 - e)/16(x - 2)#
#y - (4e^2 + e)/8 = (4e^2x - 8e^2 - ex + 2e)/16#
#y = (4e^2x - 8e^2 - ex + 2e)/16 + (4e^2 + e)/8#
#y = (4e^2x - 8e^2 - ex + 2e + 8e^2 + 2e)/16#
#y = 1/16e(4ex - x + 4)#
Thus, the equation of the tangent is #y = 1/16e(4ex - x + 4)#.

Practice exercises:

a) #y = sqrt(3x + 1)#, at #x = 5#
b) #y = e^(2x)/x^3 + e^(x^2 - 1)/x#, at #x = 1#

Hopefully this helps, and good luck!

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Answer 2

To find the equation of the tangent line of the function f(x) = e^x/x - e^(2x-3)/x^3 at x=2, we need to find the slope of the tangent line and a point on the line.

First, let's find the slope. We can do this by finding the derivative of the function f(x).

The derivative of f(x) = e^x/x - e^(2x-3)/x^3 can be found using the quotient rule.

After differentiating and simplifying, we get:

f'(x) = (x^2e^x - 3e^(2x-3) - 2xe^(2x-3))/x^4

Now, let's find the slope at x=2 by substituting x=2 into f'(x):

f'(2) = (2^2e^2 - 3e^(2(2)-3) - 2(2)e^(2(2)-3))/2^4

Simplifying further, we get:

f'(2) = (4e^2 - 3e^1 - 4e^1)/16

f'(2) = (4e^2 - 7e)/16

Now that we have the slope, let's find a point on the line. We can do this by substituting x=2 into the original function f(x):

f(2) = e^2/2 - e^(2(2)-3)/2^3

Simplifying further, we get:

f(2) = e^2/2 - e^1/8

Now we have a point (2, e^2/2 - e^1/8) on the tangent line.

Using the point-slope form of a line, we can write the equation of the tangent line:

y - (e^2/2 - e^1/8) = (4e^2 - 7e)/16 * (x - 2)

Simplifying further, we get the equation of the tangent line:

y = (4e^2 - 7e)/16 * x + (e^2/2 - e^1/8) - (4e^2 - 7e)/8

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Answer 3

To find the equation of the tangent line to ( f(x) = \frac{e^x}{x} - \frac{e^{2x-3}}{x^3} ) at ( x = 2 ), we first need to find the slope of the tangent line, which is the derivative of the function evaluated at ( x = 2 ).

[ f'(x) = \frac{d}{dx} \left( \frac{e^x}{x} - \frac{e^{2x-3}}{x^3} \right) ] [ f'(x) = \frac{e^x \cdot x - e^x - 3e^{2x-3}}{x^4} ]

Now, we can find the slope of the tangent line by evaluating ( f'(x) ) at ( x = 2 ):

[ f'(2) = \frac{e^2 \cdot 2 - e^2 - 3e^1}{2^4} ] [ f'(2) = \frac{2e^2 - e^2 - 3e}{16} ] [ f'(2) = \frac{e^2 - 3e}{16} ]

Now that we have the slope of the tangent line, we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is:

[ y - y_1 = m(x - x_1) ]

Substituting ( x_1 = 2 ), ( y_1 = f(2) ), and ( m = f'(2) ):

[ y - f(2) = \frac{e^2 - 3e}{16}(x - 2) ]

We just need to find ( f(2) ) to complete the equation.

[ f(2) = \frac{e^2}{2} - \frac{e^1}{2^3} = \frac{e^2}{2} - \frac{e}{8} ]

Thus, the equation of the tangent line at ( x = 2 ) is:

[ y - \left( \frac{e^2}{2} - \frac{e}{8} \right) = \frac{e^2 - 3e}{16}(x - 2) ]

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Answer from HIX Tutor

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