Math, asked by rushimusale3333, 9 months ago

finddy÷dx. y=log(sin(4x-3))​

Answers

Answered by SteffiPaul
0

Given,

  • y=log(sin(4x-3))​

To find,

  • We have to find dy/dx.

Solution,

We can simply find dy/dx by differentiating y with respect to x.

It is given that y = log(sin(4x-3))​

Now, differentiate both sides with respect to x, we get

                      d(y)/dx = d(log(sin(4x-3))​/dx

As we know that d(logx)/dx = 1/x, d(sinx)/dx = cosx , d(4x)/dx = 4 then

                      dy/dx  = 1/ (sin(4x-3)) * (cos(4x-3) * (4)

                       dy/dx = cos(4x-3)/sin (4x-3) * 4

As we know that cosx/sinx = cotx, then

                       dy/dx = 4 cot(4x-3)

Hence, the value of dy/dx if y= log(sin(4x-3))​ is  4 cot(4x-3).

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