if ( f(x) = 6sin^{-1}(2x) ), then ( y(x) =)\nif ( f(x) = 7cos^{-1}(3x + 11) ), then ( \frac{dy}{dx} =)\nif (…

if ( f(x) = 6sin^{-1}(2x) ), then ( y(x) =)\nif ( f(x) = 7cos^{-1}(3x + 11) ), then ( \frac{dy}{dx} =)\nif ( f(x) = 2cdot\tan^{-1}(sin(x)) ), then ( y(x) =)\nif ( f(x) = 3cot^{-1}(x^{5}) ), then ( \frac{dy}{dx} =)\nif ( f(x) = 6sec^{-1}(2ln(x)) ), then ( y(x) =)\nif ( f(x) = csc^{-1}(e^{x}) ), then ( \frac{dy}{dx} =)

if ( f(x) = 6sin^{-1}(2x) ), then ( y(x) =)\nif ( f(x) = 7cos^{-1}(3x + 11) ), then ( \frac{dy}{dx} =)\nif ( f(x) = 2cdot\tan^{-1}(sin(x)) ), then ( y(x) =)\nif ( f(x) = 3cot^{-1}(x^{5}) ), then ( \frac{dy}{dx} =)\nif ( f(x) = 6sec^{-1}(2ln(x)) ), then ( y(x) =)\nif ( f(x) = csc^{-1}(e^{x}) ), then ( \frac{dy}{dx} =)

Answer

Answer:

  1. $\frac{12}{\sqrt{1 - 4x^{2}}}$
  2. $-\frac{21}{\sqrt{1-(3x + 11)^{2}}}$
  3. $\frac{2\cos x}{1+\sin^{2}x}$
  4. $-\frac{15x^{4}}{1 + x^{10}}$
  5. $\frac{6}{x\vert2\ln x\vert\sqrt{(2\ln x)^{2}-1}}$
  6. $-\frac{e^{x}}{\vert e^{x}\vert\sqrt{e^{2x}-1}}$

Explanation:

Step1: Recall the derivative formula for (y = \sin^{-1}u)

The derivative of (y=\sin^{-1}u) with respect to (x) is (y^\prime=\frac{u^\prime}{\sqrt{1 - u^{2}}}). For (y = 6\sin^{-1}(2x)), let (u = 2x), then (u^\prime=2). So (y^\prime=6\times\frac{2}{\sqrt{1-(2x)^{2}}}=\frac{12}{\sqrt{1 - 4x^{2}}})

Step2: Recall the derivative formula for (y=\cos^{-1}u)

The derivative of (y = \cos^{-1}u) with respect to (x) is (y^\prime=-\frac{u^\prime}{\sqrt{1 - u^{2}}}). For (y = 7\cos^{-1}(3x + 11)), let (u=3x + 11), then (u^\prime = 3). So (y^\prime=7\times\left(-\frac{3}{\sqrt{1-(3x + 11)^{2}}}\right)=-\frac{21}{\sqrt{1-(3x + 11)^{2}}})

Step3: Recall the derivative formula for (y=\tan^{-1}u)

The derivative of (y=\tan^{-1}u) with respect to (x) is (y^\prime=\frac{u^\prime}{1 + u^{2}}). For (y = 2\tan^{-1}(\sin x)), let (u=\sin x), then (u^\prime=\cos x). So (y^\prime=2\times\frac{\cos x}{1+\sin^{2}x}=\frac{2\cos x}{1+\sin^{2}x})

Step4: Recall the derivative formula for (y=\cot^{-1}u)

The derivative of (y=\cot^{-1}u) with respect to (x) is (y^\prime=-\frac{u^\prime}{1 + u^{2}}). For (y = 3\cot^{-1}(x^{5})), let (u = x^{5}), then (u^\prime=5x^{4}). So (y^\prime=3\times\left(-\frac{5x^{4}}{1+(x^{5})^{2}}\right)=-\frac{15x^{4}}{1 + x^{10}})

Step5: Recall the derivative formula for (y=\sec^{-1}u)

The derivative of (y=\sec^{-1}u) with respect to (x) is (y^\prime=\frac{u^\prime}{\vert u\vert\sqrt{u^{2}-1}}). For (y = 6\sec^{-1}(2\ln x)), let (u = 2\ln x), then (u^\prime=\frac{2}{x}). So (y^\prime=6\times\frac{\frac{2}{x}}{\vert2\ln x\vert\sqrt{(2\ln x)^{2}-1}}=\frac{6}{x\vert2\ln x\vert\sqrt{(2\ln x)^{2}-1}})

Step6: Recall the derivative formula for (y=\csc^{-1}u)

The derivative of (y=\csc^{-1}u) with respect to (x) is (y^\prime=-\frac{u^\prime}{\vert u\vert\sqrt{u^{2}-1}}). For (y=\csc^{-1}(e^{x})), let (u = e^{x}), then (u^\prime=e^{x}). So (y^\prime=-\frac{e^{x}}{\vert e^{x}\vert\sqrt{e^{2x}-1}})