Question

sin(ax+b)/cos(cx+d) differentiate ​

Answer 1

Step-by-step explanation:

We have, f(x)=

cos(cx+d)

sin(ax+b)

Thus using quotient rule and chain rule simualtaneously,

f

′

(x)=

[cos(cx+d)]

2

acos(ax+b).cos(cx+d)−sin(ax+b)(−csin(cx+d))

=

cos(cx+d)

acos(ax+b)

+csin(ax+b).

cos(cx+d)

sin(cx+d)

×

cos(cx+d)

1

=acos(ax+b)sec(cx+d)+csin(ax+b)tan(cx+d)sec(cx+d)

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Answer 2

Step-by-step explanation:

f(f(x))=

cf(x)+d

af(x)+b

f(f(x))=

c

cx+d

ax+b

+d

a

cx+d

ax+b

+b

f(f(x))=

acx+bc+d

2

+dcx

a

2

x+ab+bcx+bd

=x

⇒a

2

x+ab+bcx+bd=acx

2

+bcx+d

2

x+dcx

2

Given a=−d, the above equation can also be verified

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