Mathematics - Pre-calculus 22 - Inverse Derivatives

Pre-calculus 22 - Inverse Derivatives

(10 Questions)

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Mathematics

The derivatives of arcsine, arccosine, and arctangent are discussed and applied.

y= 5 x^5 - sec^{-1} x Differentiate the given function.

y'= 25 x^4 + \frac{1}{x\sqrt{x^2-1}}

y'= 25 x^4 - \frac{1}{x\sqrt{x^2-1}}

y'= 5 x^4 + \frac{1}{\sqrt{x^2-1}}

y'= 25 x^5 - \frac{1}{\sqrt{x^2+1}}

F(x) = 2 \cos z +6 \arccos (z) Differentiate the given function.

F'(x) = \sin z +\frac{6}{\sqrt{1- z^2}}

F'(x) = 2 \sin z +\frac{6}{\sqrt{1- z^2}}

F'(x) = -2 \sin z -\frac{-6}{\sqrt{1- z^2}}

F'(x) = -2 \sin z -\frac{-1}{\sqrt{1- z^2}}

Find the derivative of f(x)=arccsc (t) - arccot(t)

-\frac {1}{t\sqrt{t^2-1}} + \frac{1}{1+t^2}

-\frac {1}{\sqrt{t^2-1}} + \frac{1}{1+t^2}

\frac {1}{t\sqrt{t^2-1}} + \frac{1}{1+t^2}

-\frac {1}{t\sqrt{t^2-1}} - \frac{1}{1+t^2}

Find the first derivative of f(x) = arctan x + x^2

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

f'(x) = \frac{(2x^3 + 2x + 1)}{1+x^2 }

f'(x) = \frac{(2x^3 + 2 )}{1+x^2 }

f'(x) = \frac{(2x^3 + x + 1)}{1+x^2 }

Find the derivative of y = 3 \arccos (x^2 + 0.5).

\frac{-3x}{\sqrt{1-(x^2+0.5)^2}}

\frac{-2x}{\sqrt{1-(x^2+0.5)^2}}

\frac{-6x}{\sqrt{1-(x^2+0.5)}}

\frac{-6x}{\sqrt{1-(x^2+0.5)^2}}

Find the derivative of y = 4 \arctan 3x^4.

\frac{48 x^3}{1+9 x^4}

\frac{48 x^6}{1+9 x^8}

\frac{48 x^3}{2+9 x^8}

\frac{48 x^3}{1+9 x^8}

Find the derivative of y= \arctan(x^2+3y)

y' = \frac {2x+3}{1+(x^2+3y)^2}

y' = \frac {2x+3}{1+(x^2+3y)}

y' = \frac {1}{1+(x^2+3y)^2}

y' = \frac {2x+3}{(x^2+3y)^2}

Differentiate the following function \arcsin 5x

y' = \frac{1}{\sqrt{1- 25x^2}}

y' = \frac{5}{\sqrt{1- 25x^2}}

y' = \frac{5}{\sqrt{1- 5x^2}}

y' = \frac{5}{\sqrt{25x^2}}

Differentiate the following function f(x)= 9 cos^{-1} x + 5 sin^{-1} x

f(x)= \frac{14}{\sqrt{1-x^2}}

f(x)= \frac{-4}{\sqrt{1-x^2}}

f(x)= \frac{4}{\sqrt{1-x^2}}

f(x)= \frac{4}{\sqrt{x^2-1}}

If \sin^{-1}(\frac{1}{2})=y, then what is the value of y?

\sin (y) = 2

y = \frac {\pi}{4}

y = \frac {\pi}{6}

y = \frac {1}{4}

by Krishna (Expert, Educator @Qalaxia)