How do you find the derivative of $f (x) = 3 arcsin (x^{2})$ $f(x) = 3 arcsin(x^2)$ ?

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Answer 1 · 2017-02-20T14:06:24

$\frac{d y}{d x} = \frac{6 x}{\sqrt{1 - x^{4}}}$ $dy/dx=(6x)/sqrt(1-x^4)$

Let $y = f (x)$ $y=f(x)$

$y = 3 arcsin (x^{2})$ $y=3arcsin(x^2)$

This suggests that $sin (\frac{y}{3}) = x^{2}$ $sin(y/3)=x^2$

Taking the derivative of both sides, we get:

$\frac{1}{3} cos (\frac{y}{3}) \frac{d y}{d x} = 2 x$ $1/3cos(y/3)dy/dx=2x$

Rearranging and cleaning it up, we get:

$\frac{d y}{d x} = \frac{6 x}{cos (\frac{y}{3})}$ $dy/dx=(6x)/(cos(y/3))$

We now need to rewrite $cos (\frac{y}{3})$ $cos(y/3)$ in terms of $x$ $x$ . We can do this using ${sin}^{2} A + {cos}^{2} A = 1$ $sin^2A+cos^2A=1$

${cos}^{2} (\frac{y}{3}) + {sin}^{2} (\frac{y}{3}) = 1$ $cos^2(y/3)+sin^2(y/3)=1$ where ${sin}^{2} (\frac{y}{3}) = x^{4}$ $sin^2(y/3)=x^4$

${cos}^{2} (\frac{y}{3}) = 1 - x^{4}$ $cos^2(y/3)=1-x^4$

$cos (\frac{y}{3}) = \sqrt{1 - x^{4}}$ $cos(y/3)=sqrt(1-x^4)$

$\frac{d y}{d x} = \frac{6 x}{\sqrt{1 - x^{4}}}$ $dy/dx=(6x)/sqrt(1-x^4)$

How do you find the derivative of f(x)=3arcsin(x2)f(x) = 3 arcsin(x^2)?