As light from a star spreads out and weakens, do gaps form between the photons?
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Answer:
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Differentiation
To Differentiate :-
f(x)=\cot x \cdot \ln \sec xf(x)=cotx⋅lnsecx
Solution :-
f(x)=\cot x \cdot \ln \sec xf(x)=cotx⋅lnsecx
\dfrac{d(f(x))}{dx}=\dfrac{d(\cot x \cdot \ln \sec x)}{dx}
dx
d(f(x))
=
dx
d(cotx⋅lnsecx)
\dfrac{d(f(x))}{dx}=\ln \sec x\cdot\dfrac{d(\cot x )}{dx}+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}
dx
d(f(x))
=lnsecx⋅
dx
d(cotx)
+cotx⋅
dx
d(lnsecx)
\left(\because \dfrac{d(fg)}{dx}=g\cdot \dfrac{d(f)}{dx}+f\cdot\dfrac{d(g)}{dx} \right)(∵
dx
d(fg)
=g⋅
dx
d(f)
+f⋅
dx
d(g)
)
\dfrac{d(f(x))}{dx}=\ln \sec x\cdot(-\csc^2x)+\cot x\cdot\dfrac{d(\ln\sec x )}{dx}
dx
d(f(x))
=lnsecx⋅(−csc
2
x)+cotx⋅
dx
d(lnsecx)
\left(\because \dfrac{d(\cot x)}{dx}=-\csc^2x \right)(∵
dx
d(cotx)
=−csc
2
x)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot\dfrac{d(\sec x )}{dx}
dx
d(f(x))
=−csc
2
x⋅lnsecx+cotx⋅
secx
1
⋅
dx
d(secx)
\left(\because \dfrac{d(\ln t)}{dx}=\dfrac{1}{t}\cdot\dfrac{dt}{dx} \right)(∵
dx
d(lnt)
=
t
1
⋅
dx
dt
)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+\cot x\cdot\dfrac{1}{\sec x}\cdot \sec x\cdot \tan x
dx
d(f(x))
=−csc
2
x⋅lnsecx+cotx⋅
secx
1
⋅secx⋅tanx
\left(\because \dfrac{d(\sec x)}{dx}=\sec x\cdot \tan x \right)(∵
dx
d(secx)
=secx⋅tanx)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+(\cot x\cdot\tan x)\cdot\left (\dfrac{1}{\cancel{\sec x}}\cdot \cancel{\sec x}\right)
dx
d(f(x))
=−csc
2
x⋅lnsecx+(cotx⋅tanx)⋅(
secx
1
⋅
secx
)
\dfrac{d(f(x))}{dx}=-\csc^2x \cdot\ln \sec x+1
dx
d(f(x))
=−csc
2
x⋅lnsecx+1
(\because \cot x \cdot \tan x = 1)(∵cotx⋅tanx=1)
Answer :-
\underline{\boxed{\dfrac{d(f(x))}{dx}=1-\csc^2x \cdot\ln \sec x}}
dx
d(f(x))
=1−csc
2
x⋅lnsecx
Note : csc x = cosec x