Question

find derivative of sin square x with respect tan x​

Answer 1

Answer:sin2x .cos^2 x

Step-by-step explanation:

Let y=sin^2 x

*differentiating with respect to x

y' = 2sin (x) cos (x) ---(1)

Let p=tan x

*differentiating w.r.t x

p' =sec^2 (x) ---(2)

(1)/(2)

dy/dp= 2sin (x) cos^3 (x)

= sin 2x cos^2 (x)

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Answer 2

First, let y=sin(x)tan(x) . Next, take the natural logarithm of both sides and use a property of logarithms to get ln(y)=tan(x)ln(sin(x)) . =1+ln(sin(x))sec2(x) . Multiplying both sides by y=sin(x)tan(x) now gives the final answer to be ddx(sin(x)tan(x))=(1+ln(sin(x))sec2(x))⋅