What is the derivative of (x^2+4x+5)/(x^3-2x)?
quotient rule !
dy/dx = ( (x^3 - 2x)(2x + 4) - (x^2 + 4x + 5)(3x^2 - 2) )/(x^3 - 2x)^2
= etc
I know you use quotient rule. Is the answer (-x^4-8x^3-17x^2+8x+10)/(x^3-2x)^2
Thanks Reiny!
If you are taking Calculus, surely you should have no difficulty expanding and simplifying
(x^3 - 2x)(2x + 4) - (x^2 + 4x + 5)(3x^2 - 2)
I will leave that up to you to check if your answer is correct.
(hint: the answer is yes)
To find the derivative of the function (x^2+4x+5)/(x^3-2x), we can use the quotient rule.
The quotient rule states that for a function f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, the derivative is given by:
f'(x) = (g'(x)*h(x) - g(x)*h'(x))/(h(x))^2
Now, let's apply the quotient rule to find the derivative of (x^2+4x+5)/(x^3-2x).
Step 1: Identify the numerator and denominator functions:
g(x) = x^2+4x+5
h(x) = x^3-2x
Step 2: Find the derivatives of the numerator and denominator functions:
g'(x) = 2x+4
h'(x) = 3x^2-2
Step 3: Plug the values into the quotient rule formula:
f'(x) = (g'(x)*h(x) - g(x)*h'(x))/(h(x))^2
= ((2x+4)*(x^3-2x) - (x^2+4x+5)*(3x^2-2))/((x^3-2x))^2
Step 4: Simplify and expand the expression:
f'(x) = (2x^4 - 4x^2 + 4x^3 - 8x - 3x^4 + 2x - 3x^2 + 2)/((x^3-2x))^2
= (-x^4 + x^3 - 7x^2 - 6x + 2)/((x^3-2x))^2
Therefore, the derivative of (x^2+4x+5)/(x^3-2x) is (-x^4 + x^3 - 7x^2 - 6x + 2)/((x^3-2x))^2.