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I have a calculus I question about derivatives:1) Find the derivative of y=cos^-1(2x)

Sagot :

[tex]y=\cos^{-1}(2x)[/tex]

1. Let u=cos(2x)

[tex]\begin{gathered} y=u^{-1} \\ \\ Chain\text{ }rule: \\ \frac{d}{dx}f(u)=\frac{d}{du}(f(u))*\frac{d}{dx}(u) \\ \\ \frac{dy}{dx}=\frac{d}{du}u^{-1}*\frac{d}{dx}cos(2x) \\ \\ \frac{dy}{dx}=-1u^{-2}*(-2sin(2x)) \\ \\ u=cos(2x) \\ \\ \frac{dy}{dx}=-cos^{-2}(2x)*(-2sin(2x)) \end{gathered}[/tex]

Simplify:

[tex]\begin{gathered} \frac{dy}{dx}=-\frac{1}{cos^2(2x)}*(-2sin(2x)) \\ \\ \frac{dy}{dx}=\frac{2sin(2x)}{cos(2x)cos2x)} \\ \\ \frac{dy}{dx}=2*\frac{sin(2x)}{cos(2x)}*\frac{1}{cos(2x)} \\ \\ \frac{sinx}{cosx}=tanx \\ \\ \frac{1}{cosx}=secx \\ \\ \\ \frac{dy}{dx}=2tan(2x)sec(2x) \end{gathered}[/tex]

Then, the derivate of the given function is:

2tan(2x)sec(2x)

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