What is the derivative of $f (x) = \frac{{sin}^{2} x + cos x}{sin x - {cos}^{2} x}$ $f(x)=(sin^2x + cosx) / (sinx - cos^2x)$ ?

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What is the derivative of $f (x) = \frac{{sin}^{2} x + cos x}{sin x - {cos}^{2} x}$ $f(x)=(sin^2x + cosx) / (sinx - cos^2x)$ ?

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Answer 1 · 2018-02-13T09:13:42

the derivative of
$\frac{{sin}^{2} x + cos x}{sin x - {cos}^{2} x}$ $(sin^2x+cosx)/(sinx-cos^2x)$ is
$\frac{sin x cos x (sin x - cos x) - (1 + 2 sin x)}{{(sin x - {cos}^{2} x)}^{2}}$ $(sinxcosx(sinx-cosx)-(1+2sinx))/(sinx-cos^2x)^2$

Explanation:

Given:
$f (x) = \frac{{sin}^{2} x + cos x}{sin x - {cos}^{2} x}$ $f(x)=(sin^2x+cosx)/(sinx-cos^2x)$
Let $u = {sin}^{2} x + cos x$ $u=sin^2x+cosx$
Then,
$\frac{d u}{d x} = 2 sin x cos x - sin x$ $(du)/dx=2sinxcosx-sinx$
Let $v = sin x - {cos}^{2} x$ $v=sinx-cos^2x$
Then,
$\frac{d v}{d x} = cos x - 2 cos x (- sin x) = cos x + 2 cos x sin x$ $(dv)/dx=cosx-2cosx(-sinx)=cosx+2cosxsinx$
Rearranging
$\frac{d v}{d x} = 2 sin x cos x + cos x$ $(dv)/dx=2sinxcosx+cosx$

We have,
$\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v^{2}}$ $d/dx(u/v)=(v(du)/dx-u(dv)/dx)/v^2$
Substituting
$\frac{d}{d x} (\frac{{sin}^{2} x + cos x}{sin x - {cos}^{2} x}) = \frac{(sin x - {cos}^{2} x) (2 sin x cos x - sin x) - ({sin}^{2} x + cos x) (2 sin x cos x + cos x)}{{(sin x - {cos}^{2} x)}^{2}}$ $d/dx((sin^2x+cosx)/(sinx-cos^2x))=((sinx-cos^2x)(2sinxcosx-sinx)-(sin^2x+cosx)(2sinxcosx+cosx))/(sinx-cos^2x)^2$

Simplifying
$\frac{(2 {sin}^{2} x cos x - 2 sin x {cos}^{3} x - {sin}^{2} x + {cos}^{2} x sin x) - (2 {sin}^{3} x cos x + 2 sin x {cos}^{2} x + {sin}^{2} x cos x + {cos}^{2} x)}{{(sin x - {cos}^{2} x)}^{2}}$ $((2sin^2xcosx-2sinxcos^3x-sin^2x+cos^2xsinx)-(2sin^3xcosx+2sinxcos^2x+sin^2xcosx+cos^2x))/(sinx-cos^2x)^2$

$\frac{(2 {sin}^{2} x cos x - 2 sin x {cos}^{3} x - {sin}^{2} x + {cos}^{2} x sin x - 2 {sin}^{3} x cos x - 2 sin x {cos}^{2} x - {sin}^{2} x cos x - {cos}^{2} x)}{{(sin x - {cos}^{2} x)}^{2}}$ $((2sin^2xcosx-2sinxcos^3x-sin^2x+cos^2xsinx-2sin^3xcosx-2sinxcos^2x-sin^2xcosx-cos^2x))/(sinx-cos^2x)^2$

$\frac{{sin}^{2} x cos x - {cos}^{2} x sin x - ({sin}^{2} x + {cos}^{2} x) - 2 sin x ({cos}^{2} x + {sin}^{2} x)}{{(sin x - {cos}^{2} x)}^{2}}$ $(sin^2xcosx-cos^2xsinx-(sin^2x+cos^2x)-2sinx(cos^2x+sin^2x))/(sinx-cos^2x)^2$

Knowing that ${cos}^{2} x + {sin}^{2} x = 1 = {sin}^{2} x + {cos}^{2} x$ $cos^2x+sin^2x=1=sin^2x+cos^2x$

$\frac{{sin}^{2} x cos x - {cos}^{2} x sin x - 1 - 2 sin x (1)}{{(sin x - {cos}^{2} x)}^{2}}$ $(sin^2xcosx-cos^2xsinx-1-2sinx(1))/(sinx-cos^2x)^2$
$\frac{sin x cos x (sin x - cos x) - 1 - 2 sin x}{{(sin x - {cos}^{2} x)}^{2}}$ $(sinxcosx(sinx-cosx)-1-2sinx)/(sinx-cos^2x)^2$
$\frac{sin x cos x (sin x - cos x) - (1 + 2 sin x)}{{(sin x - {cos}^{2} x)}^{2}}$ $(sinxcosx(sinx-cosx)-(1+2sinx))/(sinx-cos^2x)^2$
Thus, the derivative of
$\frac{{sin}^{2} x + cos x}{sin x - {cos}^{2} x}$ $(sin^2x+cosx)/(sinx-cos^2x)$ is
$\frac{sin x cos x (sin x - cos x) - (1 + 2 sin x)}{{(sin x - {cos}^{2} x)}^{2}}$ $(sinxcosx(sinx-cosx)-(1+2sinx))/(sinx-cos^2x)^2$

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What is the derivative of $f (x) = \frac{{sin}^{2} x + cos x}{sin x - {cos}^{2} x}$ $f(x)=(sin^2x + cosx) / (sinx - cos^2x)$ ?

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Explanation:

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