Question

Differentiate the following w.r.to x
log(sin/1+sinx)

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Answer 1

Answer:

Step-by-step explanation:

formula: -

given: -

y = (x)cos x + (sin x)tan x

Let u = xcosx

taking log both sides

log u = cos x. log x

now diff.w.r.to x

……………………(i)

let v = sin xtanx

taking log both sides,

log v = tan x log sin x

diff. w.r.t x, get

--------------(ii)

Adding both (i) and (ii)

= sin xtanx {1+sec2 x. log sin x}

Answer 2

Answer:

cot / 1 + sin x

Step-by-step explanation:

W.K.T

d/dx(log x) = 1/x

here x is sinx/1+sinx

So, d/dx [log(sinx/1+sinx)] is

= 1/sinx/1+sinx * (1+sinx)(cosx) - (sinx)(cosx) / $(1+sinx)^{2}$

[ we did the second part differentiation since we have x as sin / 1+sinx ]

We used the formula d/dx (u/v) = vu' - uv' / $v^{2}$

[the ' indicates differentiation ]

Here, u = sinx and v = 1+sinx

By simplifying, we get

1+sinx / sinx * cox -sinxcosx + sinx cox /  $(1+sinx)^{2}$  {+ and - gets cancelled}

We get,

1+sinx / sinx * cox /  $(1+sinx)^{2}$   { 1+sinx numerator gets cancelled to 1+sinx in denominator}

So, cosx / sinx * 1+sinx

We get,

cotx / 1+sinx

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