Find the derivative of the function.f(x)=2x−1x3+2f\left(x\right)=\frac{2x-1}{x^3+2}

− 4 x 3 + 3 x 2 + 4 ( x 3 + 2 ) 2 \frac{-4x^3+3x^2+4}{(x^3+2)^2}

Find the derivative of the function. f ( x ) = 2 x − 1 x 3 + 2 f\left(x\right)=\frac{2x-1}{x^3+2}

this problem, we're asked to find the derivative of the function f of x is equal to 2x minus 1 over x cubed plus 2. Now we have 2 functions being divided, meaning we need to go ahead and apply the quotient rule. Now in finding the derivative here, f prime of x, remember, using the quotient rule, we can think to ourselves a low d high minus high d low over the square of what's below because we have our function on the top. That is our high function, and then we have our function on the bottom, our low function. So thinking to ourselves low d high, I'm taking that bottom function there for my first term, x cubed plus 2, multiplying it by the derivative of that high function. Now the derivative of 2x minus 1 is just going to be 2. So I can move on to that next term in my numerator. I had low d high. This is high d low. We're subtracting, of course, here. So taking that high function, 2x minus 1, and multiplying it multiplying it by the derivative of that low function, that bottom function. The derivative of x cubed plus 2 is going to be 3x2. That's what I'm gonna be multiplying here. Then all of this is divided by the square of what's below. Don't forget that it's the square. So I have this x cubed plus 2 squared. Now from here, we can do some simplification. We have fully applied that quotient rule, and it is all about the algebra now. So So we can go ahead and distribute this 2 into our parentheses here as well as distribute this 3x squared. So in my numerator, I have 2x cubed plus 4 having distributed that 2 into my parentheses here. Then I am subtracting. I'm gonna keep all of this in parentheses for now to keep this organized. 3x squared times 2x is going to give us 6x cubed and then a minus 3x squared. And that is my numerator with everything multiplied out. Then my denominator is the same, x cubed plus 2 squared. Don't forget to keep squaring that every time you rewrite it. Now from here, I can distribute this negative in here. I'm just waiting to do that so that I can keep everything organized. If you could already distribute that in that single step, you can feel free to do that as well. So I have 2x cubed plus 4, then I have a minus 6x cubed plus 3x squared because minus minus gives me less. Then I'm dividing all of this by that same denominator x cubed plus 2 squared. And from here, I can combine some like terms because I have 2x cubed and I have negative 6x cubed. Now 2x cubed minus 6 x cubed is going to give us a negative 4x cubed. Then rewriting this in order so that we have a standard form polynomial, I'm gonna have plus 3x squared plus 4 with that decreasing power for standard form. Then keeping my denominator as is, of course, x cubed plus 2 squared, and we are all done here. We have found our derivative f prime of x using the quotient rule. Feel free to let me know if you have any questions here, and let's continue practicing.

Find the derivative of the function. f ( x ) = 2 x − 1 x 3 + 2 f\left(x\right)=\frac{2x-1}{x^3+2}

− 4 x 3 + 3 x 2 + 4 ( x 3 + 2 ) 2 \frac{-4x^3+3x^2+4}{(x^3+2)^2}

4 x 3 − 3 x 2 − 4 ( x 3 + 2 ) 2 \frac{4x^3-3x^2-4}{(x^3+2)^2}

4 x 3 − 3 x 2 − 4 ( 2 x − 1 ) 2 \frac{4x^3-3x^2-4}{(2x-1)^2}

8 x 3 + 3 x 2 + 4 ( x 3 + 2 ) 2 \frac{8x^3+3x^2+4}{(x^3+2)^2}

3. Techniques of Differentiation

Product and Quotient Rules

Calculus

3. Techniques of Differentiation

Product and Quotient Rules

Struggling with Calculus?

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Multiple Choice

Find the derivative of the function.
$f (x) = \frac{2 x - 1}{x^{3} + 2}$

$\frac{4 x^{3} - 3 x^{2} - 4}{(x^{3} + 2)^{2}}$

$\frac{4 x^{3} - 3 x^{2} - 4}{(2 x - 1)^{2}}$

$\frac{- 4 x^{3} + 3 x^{2} + 4}{(x^{3} + 2)^{2}}$

$\frac{8 x^{3} + 3 x^{2} + 4}{(x^{3} + 2)^{2}}$

Verified step by step guidance

Identify the function for which you need to find the derivative: $ f(x) = \frac{2x - 1}{x^3 + 2} $. This is a rational function, so you will use the quotient rule to find its derivative.

Recall the quotient rule for derivatives: If $ f(x) = \frac{u(x)}{v(x)} $, then $ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} $. Here, $ u(x) = 2x - 1 $ and $ v(x) = x^3 + 2 $.

Find the derivative of the numerator $ u(x) = 2x - 1 $. The derivative $ u'(x) $ is $ 2 $.

Find the derivative of the denominator $ v(x) = x^3 + 2 $. The derivative $ v'(x) $ is $ 3x^2 $.

Substitute $ u(x) $, $ u'(x) $, $ v(x) $, and $ v'(x) $ into the quotient rule formula: $ f'(x) = \frac{2(x^3 + 2) - (2x - 1)(3x^2)}{(x^3 + 2)^2} $. Simplify the expression to find the derivative.

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