What is the first and second derivative of 3/(1+2(x^3)) ?
im tempted to do the quotient rule, but this looks weird to me...any advice would be greatly appreciated!!
let u=(1+2x^3)
then f= 3u^-1
f'= -3 *u^-2 *u'
but u'= 6x^2
so f'= -3*6x^2 /(1+2x^3)^2
In google type: calc101
When you see list of resultc click on:
Calc101com Automatic Calculus,Linear Algebra and Polynomials
When page be open clik option: derivatives
When this page be open in rectacangle type:
3/(1+2(x^3))
and click options DO IT
You will see solution step-by-step
By the way on this site you can practice any kind of derivation.
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To find the first and second derivative of the function f(x) = 3/(1 + 2x^3), we can use the quotient rule.
The quotient rule states that for a function u/v, where u and v are both differentiable functions, the first derivative can be calculated as:
f'(x) = (v * u' - u * v') / v^2,
where u' represents the derivative of u and v' represents the derivative of v.
Let's apply the quotient rule to find the first derivative:
First, let u = 3 and v = 1 + 2x^3.
Differentiating u with respect to x, we get u' = 0 (since 3 is a constant).
Differentiating v with respect to x, we get v' = 6x^2.
Applying the quotient rule, we have:
f'(x) = ((1 + 2x^3) * 0 - 3 * 6x^2) / (1 + 2x^3)^2.
Simplifying this expression, we get:
f'(x) = -18x^2 / (1 + 2x^3)^2.
This is the first derivative of the given function. Now, let's find the second derivative.
To find the second derivative, we differentiate f'(x) with respect to x:
f''(x) = d/dx (-18x^2 / (1 + 2x^3)^2).
We can use the quotient rule again to differentiate this expression:
f''(x) = (0 - (-18x^2) * (2(1 + 2x^3)^2)' ) / (1 + 2x^3)^4.
Simplifying this, we get:
f''(x) = (36x^2 * (1 + 6x^3)) / (1 + 2x^3)^4.
Therefore, the first derivative of 3/(1 + 2(x^3)) is -18x^2 / (1 + 2x^3)^2, and the second derivative is (36x^2 * (1 + 6x^3)) / (1 + 2x^3)^4.