Implicit Differentiation - Inverse Trigonometric Functions Practice Problems Online

If $\displaystyle \tan^{-1} \left(x^2y\right)=3x+6y,$ what is $\frac{dy}{dx} ?$

\displaystyle{y'=\frac{6+6x^4y^2-2xy}{x^2-3-3x^4y^2}}

\displaystyle{y'=\frac{3+3x^4y^2-2xy}{x^2-6-6x^4y^2}}

\displaystyle{y'=\frac{6+3x^4y^2-2xy}{x^2-6-3x^4y^2}}

\displaystyle{y'=\frac{3+6x^4y^2-2xy}{x^2-3-6x^4y^2}}

Given the function $\tan^{-1}(4xy)=5,$ What is $\frac{dy}{dx}?$

If $\displaystyle \cos^{-1} \left(\frac{6x-y}{3x+y}\right)=e^a,$ what is $\frac{dy}{dx} ?$

\displaystyle{ \frac{dy}{dx}=\frac{y}{x}}

\displaystyle{ \frac{dy}{dx}=\frac{x^2}{y^2}}

\displaystyle{ \frac{dy}{dx}=\frac{x}{y}}

\displaystyle{ \frac{dy}{dx}=\frac{y^2}{x^2}}

If $\displaystyle \sin^{-1} (7x^2-y^2)=\ln 6,$ what is $\frac{dy}{dx} ?$

Given the function $\cos^{-1}(3x+11y)=12,$ what is $\frac{dy}{dx} ?$

\frac{dy}{dx}=\frac{11}{12}

\frac{dy}{dx}=-\frac{12}{11}

\frac{dy}{dx}=\frac{11}{3}

\frac{dy}{dx}=-\frac{3}{11}

Implicit Differentiation - Inverse Trigonometric Functions