Question

Question #48c60

Answer 1

`-(2sec^2x)/(1+tanx)^2`

Explanation:

differentiate using the `color(blue)"quotient rule"`

`"Given " f(x)=(g(x))/(h(x))" then"`

`• f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2`

`color(orange)"Reminder" d/dx(tanx)=sec^2x`

`"here " g(x)=1-tanxrArrg'(x)=-sec^2x`

`"and " h(x)=1+tanxrArrh'(x)=sec^2x`

`rArrf'(x)=((1+tanx)(-sec^2x)-(1-tanx)(sec^2x))/(1+tanx)^2`

`color(white)(rArrf'(x))=(-sec^2xcancel(-sec^2xtanx)-sec^2xcancel(+sec^2xtanx))/(1+tanx)^2`

`color(white)(rArrf'(x))=-(2sec^2x)/(1+tanx)^2`

Answer 2

`-2/(cosx+sinx)^2,`
or,
`-2/(1+sin2x).`

Explanation:

Observe that, `(1-tanx)/(1+tanx)=tan(pi/4-x).`

Hence, `d/dx{(1-tanx)/(1+tanx)}=d/dx{tan(pi/4-x),`

`=sec^2(pi/4-x)*d/dx(pi/4-x),...[because," The Chain Rule]"`

`=-sec^2(pi/4-x),`

`=-1/{cos(pi/4-x)}^2,`

`-1/{cos(pi/4)cosx+sin(pi/4)sinx}^2,`

`=-1/{1/sqrt2(cosx+sinx)}^2,`

`=-2/(cosx+sinx)^2, or,`

`=-2/{cos^2x+2cosxsinx+sin^2x},`

`=-2/(1+sin2x).`

Enjot Maths.!