What is the derivative of $f (x) = 3 arcsin (x^4)$ ?

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Answer 1 · 2016-07-09T04:56:42

$f'(x)=(12x^3)/{sqrt(1-x^8)}.$

Let $y=f(x)=3arcsinx^4.$

We know that $d/dt(arcsint)=1/sqrt(1-t^2).$

Let, $y=3arcsint$ , where, $t=x^4.$ Thus, $y$ becomes a fun. of $t$ , and, $t$ of $x$ .

Therefore, by Chain Rule, $dy/dx=dy/dt*dt/dx.$

$=d/dt(3arcsint)*d/dx(x^4),$

$=3*d/dt(arcsint)*4*x^(4-1),$
$=12*1/sqrt(1-t^2)*x^3,$
$=(12x^3)/[sqrt{1-(x^4)^2)},$
$:. f'(x)=(12x^3)/{sqrt(1-x^8)}.$

Enjoy Maths.!

What is the derivative of f (x) = 3 arcsin (x^4)f (x) = 3 arcsin (x^4)?