Question

f(x)=4arcsin(x^3), find f'(x). derivative

Preferred: looking for paper picture on how to do it with written steps.

Answer 1

do you mean find \(f'(x)\) ?

Answer 2

oh yes, derivative, mistake

Answer 3

\[\sin\frac{f(x)}{4}=x^3\]
\[\frac{d}{dx}\sin\frac{f(x)}{4}=\frac{d}{dx}x^3\]Chain rule:
\[\frac{1}{4}\cos\frac{f(x)}{4}\frac{df(x)}{dx}=3x^2\]
\[\large \frac{df(x)}{dx}=12\frac{x^2}{\cos\frac{f(x)}{4}}=12\frac{x^2}{\cos arcsin(x^3)}\]

Answer 4

looking for another way of doing this problem

Answer 5

f(x)=4arcsin(x^(3))

The derivative of f(x) is equal to f'(x)=(d)/(dx) 4arcsin(x^(3)).
f'(x)=(d)/(dx) 4arcsin(x^(3))

Find the derivative of the expression.
(d)/(dx) 4arcsin(x^(3))

The chain rule states that the derivative of a composite function (f o g)' is equal to (f' o g)*g'. To find the derivative of 4arcsin(x^(3)), find the derivatives of each portion of the function and use the chain rule formula.
(d)/(du) 4arcsin(u)*(d)/(dx) x^(3)

The derivative of 4arcsin(u) is 4((1)/(sqrt(1-u^(2)))).
(d)/(du) 4arcsin(u)=4((1)/(sqrt(1-u^(2))))

Multiply 4 by each term inside the parentheses.
(d)/(du) 4arcsin(u)=(4)/(sqrt(1-u^(2)))

To find the derivative of x^(3), multiply the base (x) by the exponent (3), then subtract 1 from the exponent.
(d)/(dx) x^(3)=3x^(2)

Replace the variable u with x^(3) in the expression.
(d)/(du) 4arcsin(u)=(4)/(sqrt(1-(x^(3))^(2)))

Replace the variable u with x^(3) in the expression.
3x^(2)

Form the derivative by substituting the values for each portion into the chain rule formula.
=(4)/(sqrt(1-(x^(3))^(2)))*3x^(2)

Multiply 4 by 3x^(2) to get 12x^(2).
(d)/(dx) 4arcsin(x^(3))=(12x^(2))/(sqrt(1-(x^(3))^(2)))

The derivative of the function is (12x^(2))/(sqrt(1-(x^(3))^(2))).
f'(x)=(12x^(2))/(sqrt(1-(x^(3))^(2)))