Differentiate f (x) = sin^-1(x^4) Find the value of the derivative at x=1/3?

f(x) = sin⁻¹(x⁴)

f '(x) = d sin⁻¹(x⁴) / d x
= [d sin⁻¹(x⁴) / d x⁴] * (d x⁴ / d x)
= {1 / √[1 - (x⁴)²]} * (4x³)
= 4x³ / √(1 - x⁸)

The derivative at x = 1/3, f' (1/3)
= 4(1/3)³ / √[1 - (1/3)⁸]
= (4/27) / √(6560/6561)
= (4/27) / [4(√410)/81]
= 3/√410
= 3(√410)/410

https://c2.staticflickr.com/8/7504/26912017436_bcda38d530_o.png

f(x) = sinˉ¹(x^4)
f'(x) = 1/√[1 - (x^4)²] * 4x³
f'(x) = 4x³/ √(1 - x^8)

f'(1/3) = 4(1/3)³ / √[1 - (1/3)^8)
f'(1/3) = (4/27) / √[1 - (1/6561)]
f'(1/3) = (4/27) / (√6560/6561)
f'(1/3) = 4 / 27(√6560/6561)
f'(1/3) = 0.148

u = x^4 .... du = 4x^3
f ' = du / sqrt(1 - u^2)
f '(x) = 4x^3 / sqrt(1 - x^8)

f '(1/3) = (4/27) / sqrt(1 - 1/3^8) <<< simplify as needed or use a calculator

The derivative of sin^-1 (x) is 1 / sqrt(1 - x^2)

so...

[ 1 / sqrt(1 - x^2) ] * chain rule
[ 1 / sqrt(1 - x^2) ] * 4x^3
4x^3 / sqrt(1 - x^2)

I'm sure you can plug in 1/3 and get your answer just fine
I got .1571 ...