Question

Найти производную обратной функции.
f(x)=arcsin3x
f(x)=x^3-arccos2x

Answer 1

Ответ:

$2a)\ \ y=arcsin3x\ \ \to \ \ 3x=siny\ \ ,\ \ x=\varphi (y)=\dfrac{1}{3}\, siny\\\\f'(x)=\dfrac{1}{\varphi'(y)}\ \ \to \ \ \varphi'(y)=\dfrac{1}{f'(x)}=\dfrac{1}{\frac{3}{\sqrt{1-9x^2}}}=\dfrac{\sqrt{1-9x^2}}{3}=\dfrac{\sqrt{1-9\cdot \frac{1}{9}sin^2y}}{3}=\\\\=\dfrac{\sqrt{1-sin^2y}}{3\frac{x}{y} }=\dfrac{cosy}{3}=\dfrac{1}{3}\, cosy$

$P.S.\ \ \ (arcsin3x)'=f'(x)=\dfrac{1}{\varphi '(y)}=\dfrac{1}{\frac{1}{3}\, cosy}=\dfrac{3}{\sqrt{1-sin^2y}}=\\\\\\=\dfrac{3}{\sqrt{1-sin^2(arcsin3x)}}=\dfrac{3}{\sqrt{1-(3x)^2}}=\dfrac{3}{\sqrt{1-9x^2}}$

$b)\ \ f(x)=x^3-arccos2x\\\\f'(x)=3x^2+\dfrac{2}{\sqrt{1-4x^2}}\\\\\varphi '(y)=\dfrac{1}{f'(x)}=\dfrac{1}{3x^2+\frac{2}{\sqrt{1-4x^2}}}=\dfrac{\sqrt{1-4x^2}}{3x^2\sqrt{1-4x^2}+2}$