Differential coefficient of secx with respect to tanx
Answers
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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