How do you use the chain rule to differentiate log_(13)cscx?

Aug 12, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\ln} 13 \cdot \cot x$

Explanation:

Here ,

$y = {\log}_{13} \csc x$

Using Change of Base Formula:

color(blue)(log_aX=log_k X/log_ka ,where,k " is the new base"

So .

$y = {\log}_{e} \csc \frac{x}{\log} _ e 13 = \frac{1}{\ln} 13 \left(\ln \csc x\right)$

Let ,

$y = \frac{1}{\ln} 13 \ln u \mathmr{and} u = \csc x$

$\frac{\mathrm{dy}}{\mathrm{du}} = \frac{1}{\ln} 13 \cdot \frac{1}{u} \mathmr{and} \frac{\mathrm{du}}{\mathrm{dx}} = - \csc x \cot x$

Using Chain Rule:

color(red)((dy)/(dx)=(dy)/(du)*(du)/(dx

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} 13 \cdot \frac{1}{u} \left(- \csc x \cot x\right)$

Subst. $u = \csc x$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\ln} 13 \cdot \frac{1}{\csc x} \left(- \csc x \cot x\right)$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{\ln} 13 \cdot \cot x$

Aug 14, 2018

Same as above answer, different method.

Explanation:

Alternatively...

Let $y = {\log}_{13} \left(\csc x\right)$
$\Rightarrow y = - {\log}_{13} \left(\sin x\right)$
$\Rightarrow \sin x = {13}^{-} y$

Using implicit differentiation:
$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \frac{d}{\mathrm{dx}} \left({13}^{-} y\right)$
$\Rightarrow \frac{d}{\mathrm{dx}} \left(\sin x\right) = \frac{d}{\mathrm{dy}} \left({e}^{- y \cdot \ln 13}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
$\therefore \cos x = - \ln 13 \cdot {13}^{-} y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

But ${13}^{-} y = \sin x$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \cos \frac{x}{\sin} x \div - \ln 13$
$= - \cot \frac{x}{\ln} 13$

As above.