If f(x)=5x^3–2/x^4
find f'(x)
I don't understand what I did wrong for this
((x^4)(4x+7)-(4x^3)(5x^3-2))/(x^4)^2
Now reduce the x powers. I see righ toff you can remove an x^3 power.
(4x^2+7x-12)/x^8 ??
No
Let me guide you through the u/v rule.
d(u/v)=( udv + vdu ) /v^2
u= 5x^2-2 du=10x
v=x^4 dv=4x^3
f'= (5x^2+2)x^4 + 10x(4x^3) / x^8
= ((5x^2+2) + 40))/x^4
no gather the terms, and do the quadratic. Check my typing
d/dx (5x^3–2) = 15 x^2
not what you wrote
To find the derivative of the function f(x)=5x^3–2/x^4, you can use the quotient rule. The quotient rule states that if you have a function of the form h(x) = g(x)/k(x), then the derivative h'(x) is given by:
h'(x) = (g'(x) * k(x) - g(x) * k'(x))/[k(x)]^2
Let's apply the quotient rule to find f'(x) step by step:
Step 1: Identify the functions g(x) and k(x).
In this case, g(x) = 5x^3 – 2, and k(x) = x^4.
Step 2: Find the derivative g'(x) and k'(x).
To find g'(x), take the derivative of 5x^3 – 2 with respect to x:
g'(x) = 3 * 5x^(3-1) = 15x^2
To find k'(x), take the derivative of x^4 with respect to x:
k'(x) = 4 * x^(4-1) = 4x^3
Step 3: Substitute g(x), k(x), g'(x), and k'(x) into the quotient rule formula.
f'(x) = (15x^2 * x^4 - (5x^3 – 2) * 4x^3)/(x^4)^2
Simplifying further:
f'(x) = (15x^6 - (20x^6 - 8x^3))/(x^8)
f'(x) = (15x^6 - 20x^6 + 8x^3)/(x^8)
f'(x) = (-5x^6 + 8x^3)/(x^8)
Therefore, the derivative of f(x)=5x^3–2/x^4 is f'(x) = (-5x^6 + 8x^3)/(x^8).