solve for quotient rule

Question

Find the derivative and simplify your answer. Any negative exponent in your final answer must be made positive and any rational exponent in your final answer must be expressed in radical form.

f (x) = 2 x^x - 3x+1

__________

4x+1

Answer 1

cute... but easier said than done...

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First we need Dx [x^x]

let y = x^x

ln y = x ln x

y'/y = x(1/x) + lnx

y' = (1 + xln x) y

dy = (1 + x ln x ) x^x dx = Dx[x^x]

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Now we can appeal to the quotient rule...

(4x+1) [ 2 (1 + x ln x ) x^x - 3 ] - 4 (2 x^x - 3x + 1)

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(4x+1)^2

(4x+1) [ 2x^x + 2x^(x+1) lnx - 3] - 4( 2x^x - 3x + 1)

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(4x+1)^2

Answer 2

Your function is f(x) = (2x^x - 3x + 1)/(4x+1)

This function would be quite easy to differentiate if it wasn't for the x^x, so let's take a look at what to do with that first.

We'll start with the function f(x) = x^x

This function doesn't fall nicely into any of the familiar derivative rules. This isn't an xⁿ or a a^x function. Likewise, we also can't view it as a composite function and use chain rule. Is the exponent or base the outside function?

Instead we turn to a little algebra trick using an old logarithm property: log_b(x ^y) = y ∙ log_b(x)

f(x) = x^x

ln f(x) = ln x^x (you can use any log base you want in this step)

ln f(x) = x*ln x

ln y = x*ln x (changing f(x) into y to make things simpler)

At this point it is easy to take a derivative of the right side but now that the left side is now a function of y, we have to use implicit differentiation to solve.

y'/y = ln x + x*(1/x)

y'/y = ln x + 1

y' = f'(x) = (ln x + 1)*y

f'(x) = (ln x + 1)*(x^x)

Ok, let's try the actual problem now that we know how to deal with an x^x term.

f(x) = (2x^x - 3x + 1)/(4x+1)

Our first step is to apply the quotient rule and that won't work unless we can differentiate

2x^x -3x+1

Let's apply what we learned up above.

y = 2x^x -3x+1

dy/dx = 2*dy/dx (x^x) - 3 dy/dx (x) + dy/dx (1)

dy/dx = 2[(ln x + 1)*(x^x)] - 3(1) + (0)

y' = 2[(ln x + 1)*(x^x)] - 3

Alright, I think we can finally tackle this problem using the quotient rule.

f(x) = (2x^x - 3x + 1)/(4x+1)

f'(x) = [2x^x[(ln x + 1)] - 3 ]*(4x+1) - (2x^x - 3x + 1)*(4)

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(4x+1)²

The last step is to expand and simplify. If you have any further questions, feel free to ask!

Answer 3

Hi, Ana:

Let's do this one!

First, see that we have a quotient, so will need the quotient rule.

Second, there is that tricky x^x. Let's do it separately: Let y = x^x

Now, take natural logs on both sides: ln(y) = x ln(x)

Differentiate implicitly from left to right: (1/y) y' = ln(x) + x/x

(1/y)y' = ln(x) + 1

y' = y(ln(x) + 1)

= x^x ln(x) + 1

Now, by the quotient rule: f'(x) = {(2 (x^x ln(x) + 1) - 3) (4x + 1) - 4(2x^x - 3x +1)} / (4x + 1)²