Question

differentiatie (sinx.cos^2 x + cosx. sin^2 x).

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Answer 1

Answer:

f(x)=sinx / cos2x

f'(x)=( (sinx)’*cos2x -(cos2x)’*sinx)/(cos2x)^2

= (cosxcos2x+2sin2xsinx)/(cos2x)^2

=(cosxcos2x+sinxsin2x+sinxsin2x)/(cos2x)^2

=(cos(2x-x)+sinxsin2x)/(cos2x)^2

=(cosx + sinxsin2x)/(cos2x)^2

=(cosx+2cosx(sinx)^2)/(cos2x)^2

=cosx *(1+2(sinx)^2)/(cos2x)^2

=cosx *(1+1-cos2x)/(cos2x)^2

=cosx*(2-cos2x)/(cos2x)^2

Let,

y= sinx/cos2x

y=sinx.sec2x

Now diff. w.r.t.x we get

dy/dx= d/dx[sinx.sec2x]

dy/dx= sinx.d/dx(sec2x)+ sec2x.d/dx(sinx)

dy/dx= sinx.sec2x.tan2x.d/dx(2x)+sec2x.cosx

dy/dx= sinx.sec2x.tan2x.(2)+sec2x.cosx

dy/dx= 2sinx.sec2x.tan2x+ sec2x.cosx

dy/dx= 2.sinx.sec²2x.sin2x+sec2x.cosx

➡️dy/dx=sec²2x[2sinx.sin2x+cos2x.cosx

➡️dy/dx=sec²2x.[sinx.sin2x+(cos2x.cosx+sin2x.sinx)]

➡️dy/dx= sec²2x[sinx.sin2x+cos(2x-x)]

➡️dy/dx= sec²2x[sinx.sin2x+cosx]

➡️dy/dx= sec²2x[(1/2).2sin2x.sinx+cosx]

➡️dy/dx= (1/2)sec²2x [cosx-cos3x+2cosx]

➡️dy/dx= (1/2)sec²2x[3cosx-cos3x] Ans.

Use the derivative of a quotient rule: d/dx (u/v) = (vu’-uv’)v^2

d/dx(sin x/cos2x)=cos 2x*cosx-sinx(-2sin2x)/cos^2(2x)

hope this was helpful!!

=(cos2x*cosx+2sinx2x)/cos^2(2x)