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

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

We must differentiate the numerator using the chain rule--

As for the denominator:

So, we have

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

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.

Practice exercises:

Hopefully this helps, and good luck!

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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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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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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.

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