Question

log (sin e×) ka differentiation
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Answer 1

Answer:

1/sin e^x d/dx(sin e^x)

=1/sin e^x.cos e^x d/dx(e^x)

=cose^x/sine^x.e^x

=e^x.tane^x

Answer 2

Answer:

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Step-by-step explanation:

Given that y=log(sinx), we need to find the value of

dx

dy

We know that the derivative of log(x) w.r.t x is

x

1

⟹

dx

dy

=

dx

d

(log(sinx))=

sinx

1

×

dx

d

(sinx)=

sinx

1

×cosx=cotx

Hence,

dx

dy

=cotx