Question

derivative of secx with respect to tanx is?

Answer 1

Answer:

$\displaystyle\bf\frac{d(sec\;x)}{d(tan\;x)}=sin\;x$

Step-by-step explanation:

$Let\;\;u=sec\;x\;\&\;\;v=tan\;x$

$u=sec\;x$

$\frac{du}{dx}=sec\;x\;tan\;x$

and

$v=tan\;x$

$\frac{dv}{dx}=sec^2\;x$

Now

$\displaystyle\frac{d(sec\;x)}{d(tan\;x)}$

$\displaystyle\frac{du}{dv}=\frac{\frac{du}{dx}}{\frac{dv}{dx}}$

$\displaystyle\frac{dy}{dx}=\frac{sec\;x\;tan\;x}{sec^2\;x}$

$\displaystyle\frac{dy}{dx}=\frac{tan\;x}{sec\;x}$

$\displaystyle\frac{dy}{dx}=\frac{\frac{sin\;x}{cos\;x}}{\frac{1}{cos\;x}}$

$\implies\displaystyle\boxed{\bf\frac{d(sec\;x)}{d(tan\;x)}=sin\;x}$

Answer 2

Answer:

tanx. secx

Step-by-step explanation:

secx=1/cosx

formula =f(x) /g(x) =g(x) d/dx.f(x) -f(x) d/dx.g(x)

___________________

(g(x)) ^2

=cosx.d/dx.(1)-1d/dx.cosx

_________________

(cosx)^2

=cosx(0)-1.(-sinx)

___________

: sinx/cosx=tanx cos^2x

=sinx.1

____

cos^2x

:1/cosx= secx =sinx/cosx ×1/cosx

=tanx.secx

hence, proved