What is the derivative of (sinx)^(cos^(-1)x)?

Jul 4, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\sin x\right)}^{{\cos}^{- 1} x} \setminus \left({\cos}^{- 1} x \setminus \cot x - \frac{\ln \sin x}{\sqrt{1 - {x}^{2}}}\right)$

Explanation:

Let:

$y = {\left(\sin x\right)}^{{\cos}^{- 1} x}$

Take Natural logarithms of both sides:

$\ln y = \ln \left\{{\left(\sin x\right)}^{{\cos}^{- 1} x}\right\}$
$\text{ } = \left({\cos}^{- 1} x\right) \setminus \ln \left\{\sin x\right\}$

Differentiate Implicitly,applying the product rule and chain rule:

$\frac{1}{y} \setminus \frac{\mathrm{dy}}{\mathrm{dx}} = \left({\cos}^{- 1} x\right) \left(\frac{d}{\mathrm{dx}} \ln \left\{\sin x\right\}\right) + \left(\frac{d}{\mathrm{dx}} {\cos}^{- 1} x\right) \left(\ln \left\{\sin x\right\}\right)$
$\text{ } = \left({\cos}^{- 1} x\right) \left(\frac{1}{\sin} x \setminus \cos x\right) + \left(- \frac{1}{\sqrt{1 - {x}^{2}}}\right) \left(\ln \left\{\sin x\right\}\right)$
$\text{ } = {\cos}^{- 1} x \setminus \cos \frac{x}{\sin} x - \frac{1}{\sqrt{1 - {x}^{2}}} \left(\ln \left\{\sin x\right\}\right)$
$\text{ } = {\cos}^{- 1} x \setminus \cot x - \frac{\ln \sin x}{\sqrt{1 - {x}^{2}}}$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = y \setminus \left({\cos}^{- 1} x \setminus \cot x - \frac{\ln \sin x}{\sqrt{1 - {x}^{2}}}\right)$
$\text{ } = {\left(\sin x\right)}^{{\cos}^{- 1} x} \setminus \left({\cos}^{- 1} x \setminus \cot x - \frac{\ln \sin x}{\sqrt{1 - {x}^{2}}}\right)$

Jul 4, 2017

$\frac{{\left(\sin x\right)}^{{\cos}^{- 1} x} \left\{\sqrt{1 - {x}^{2}} \cot x {\cos}^{- 1} x - \ln \left(\sin x\right)\right\}}{\sqrt{1 - {x}^{2}}} .$

Explanation:

Let, $y = {\left(\sin x\right)}^{{\cos}^{- 1} x} .$

$\therefore \ln y = \ln \left\{{\left(\sin x\right)}^{{\cos}^{-} 1 x}\right\} = \left({\cos}^{- 1} x\right) \left(\ln \left(\sin x\right)\right) .$

$\therefore \frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{d}{\mathrm{dx}} \left\{\left({\cos}^{- 1} x\right) \left(\ln \left(\sin x\right)\right)\right\} \ldots \ldots \ldots \ldots \left(\ast\right) .$

Here, by the Chain Rule,

$\frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{d}{\mathrm{dy}} \left(\ln y\right) \cdot \frac{d}{\mathrm{dx}} \left(y\right) = \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} \ldots \ldots \left({\ast}^{1}\right) .$

For the Derivative on the R.H.S., we use the Product Rule

and the Chain Rule.

$\frac{d}{\mathrm{dx}} \left\{\left({\cos}^{- 1} x\right) \left(\ln \left(\sin x\right)\right)\right\} ,$

$= {\cos}^{- 1} x \cdot \frac{d}{\mathrm{dx}} \left(\ln \left(\sin x\right)\right) + \ln \left(\sin x\right) \cdot \frac{d}{\mathrm{dx}} {\cos}^{- 1} x ,$

$= \left({\cos}^{- 1} x\right) \left\{\frac{1}{\sin} x \cdot \frac{d}{\mathrm{dx}} \left(\sin x\right)\right\}$
$+ \ln \left(\sin x\right) \cdot \left\{- \frac{1}{\sqrt{1 - {x}^{2}}}\right\} ,$

$= \left({\cos}^{- 1} x\right) \left(\cos \frac{x}{\sin} x\right) - \frac{\ln \left(\sin x\right)}{\sqrt{1 - {x}^{2}}} ,$

$= \cot x {\cos}^{- 1} x - \ln \frac{\sin x}{\sqrt{1 - {x}^{2}}} ,$

$= \frac{\sqrt{1 - {x}^{2}} \cot x {\cos}^{- 1} x - \ln \left(\sin x\right)}{\sqrt{1 - {x}^{2}}} \ldots . . \left({\ast}^{2}\right) .$

Using $\left({\ast}^{1}\right) \mathmr{and} \left({\ast}^{2}\right) \text{ in } \left(\ast\right) ,$ we have,

$\frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\sqrt{1 - {x}^{2}} \cot x {\cos}^{- 1} x - \ln \left(\sin x\right)}{\sqrt{1 - {x}^{2}}} ,$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y \left\{\sqrt{1 - {x}^{2}} \cot x {\cos}^{- 1} x - \ln \left(\sin x\right)\right\}}{\sqrt{1 - {x}^{2}}} , \mathmr{and} ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{\left(\sin x\right)}^{{\cos}^{- 1} x} \left\{\sqrt{1 - {x}^{2}} \cot x {\cos}^{- 1} x - \ln \left(\sin x\right)\right\}}{\sqrt{1 - {x}^{2}}} .$

Enjoy Maths.!