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Dafna11 [192]
2 years ago
15

Use the inverse function-inverse cofunction

Mathematics
1 answer:
vivado [14]2 years ago
4 0

Answer:

arccot(x)'=-\frac{1}{1+x^2}

Step-by-step explanation:

<u>Use implicit differentiation:</u>

y=arccot(x)

cot(y)=x

\frac{1}{tan(y)}=x

\frac{cos(y)}{sin(y)}=x

\frac{dy}{dx}(\frac{cos(y)}{sin(y)})=\frac{dy}{dx}(x)

\frac{sin(y)(-sin(y))-cos(y)cos(y)}{sin^2(y)}*\frac{dy}{dx}=1 (Quotient Rule: \frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2} )

\frac{-sin^2(y)-cos^2(y)}{sin^2(y)}*\frac{dy}{dx}=1

\frac{-(sin^2(y)+cos^2(y))}{sin^2(y)}*\frac{dy}{dx}=1

\frac{-1}{sin^2(y)}*\frac{dy}{dx}=1

-csc^2(y)\frac{dy}{dx}=1

\frac{dy}{dx}=\frac{1}{-csc^2(y)}

\frac{dy}{dx}=-sin^2(y)

\frac{dy}{dx}=-sin^2(arccot(x))

See the attached picture to understand how to evaluate the mixed composition of trig functions using a right triangle.

Therefore, the derivative of arccot(x) is -\frac{1}{\sqrt{1+x^2}}.

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