Question

Differential coefficient of secx with respect to tanx

Answer 1

Answer:

:

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

d(tanx)

d(secx)

=sinx

Step-by-step explanation:

Let\;\;u=sec\;x\;\&\;\;v=tan\;xLetu=secx&v=tanx

u=sec\;xu=secx

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

dx

du

=secxtanx

and

v=tan\;xv=tanx

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

dx

dv

=sec

2

x

Now

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

d(tanx)

d(secx)

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

dv

du

=

dx

dv

dx

du

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

dx

dy

=

sec

2

x

secxtanx

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

dx

dy

=

secx

tanx

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

dx

dy

=

cosx

1

cosx

sinx

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

d(tanx)

d(secx)

=sinx

Step-by-step explanation:

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