Question

what is the differentiation (or derivative) of
1 / (sinx + cosx)​

Answer 1

GIVEN :–

• A function $\: \: { \bold{ \dfrac{1}{ \sin(x) + \cos(x)}}}$

TO FIND :–

$\:\:\: { \bold { \dfrac{dy}{dx} = ?}}$

SOLUTION :–

• Let the function –

$\\ \implies { \bold{y = \dfrac{1}{ \sin(x) + \cos(x)}}}$

• We know that –

$\\ \implies { \bold{ \frac{d \left( \dfrac{u}{v} \right) }{dx} = \dfrac{v. \dfrac{du}{dx} - u \dfrac{dv}{dx}}{ {v}^{2} }}}$

• So that –

$\\ \implies { \bold{ \dfrac{dy}{dx} = \dfrac{ \{\sin(x) + \cos(x) \}(0) - (1) \{ \cos(x) - \sin(x) \} }{ \{ \sin(x) + \cos(x) \}^{2} }}}$

$\\ \implies { \bold{ \dfrac{dy}{dx} = \dfrac{ - (1) \{ \cos(x) - \sin(x) \} }{ \{ \sin(x) + \cos(x) \}^{2} }}}$

$\\ \implies { \bold{ \dfrac{dy}{dx} = \dfrac{ \sin(x) - \cos(x) }{ \{ \sin(x) + \cos(x) \}^{2} }}}$

$\\ \implies { \bold{ \dfrac{dy}{dx} = \dfrac{ \sin(x) - \cos(x) }{ \sin^{2} (x) + \cos^{2} (x) + 2 \sin(x) \cos(x) }}}$

$\\ \implies \large{ \boxed { \bold{ \dfrac{dy}{dx} = \dfrac{ \sin(x) - \cos(x) }{ 1 + \sin(2x)}}}}$

▪︎ OTHER USED IDENTITIES :–

$\\ { \bold{ (1) \: \: \dfrac{d( \sin x) }{dx} = \cos(x) }}$

$\\ { \bold{ (2) \: \: \dfrac{d( \cos x) }{dx} = - \sin(x) }}$

$\\ { \bold{ (3) \: \: \sin^{2} (x) + { \cos}^{2}(x) = 1 }}$

$\\ { \bold{ (4) \: \: 2 \sin(x) \cos(x) = \sin(2x) }}$

Answer 2

Given ,

The function is 1/sin(x) + cos(x)

Differentiating with respect to x , we get

$\sf \mapsto \frac{dy}{dx} = \frac{1}{ \sin(x) + \cos(x) } \\ \\\sf \mapsto \frac{dy}{dx} = \frac{\sin(x) + \cos(x) \times \frac{d(1)}{dx} - 1 \times \frac{d \{\sin(x) + \cos(x) \}}{dx} }{ ({\sin(x) + \cos(x) )}^{2} } \\ \\ \sf \mapsto \frac{dy}{dx} = \frac{ - \cos(x) + \sin(x) }{ { \sin}^{2}(x) + { \cos}^{2}(x) + 2 \sin(x) \cos(x) } \\ \\\sf \mapsto \frac{dy}{dx} = \frac{ \sin(x)- \cos(x) }{1 + \sin(2x) }$

Remmember :

$\sf \mapsto \frac{d(u.v)}{dx} = \frac{v \frac{d(u)}{dx} - u \frac{d(v)}{dx} }{ {(v)}^{2} } \\ \\\sf \mapsto \frac{d(Constant)}{dx} = 0 \\ \\ \sf \mapsto \frac{d \{ \sin(x) \}}{dx} = \cos(x) \\ \\ \sf \mapsto \frac{d \{ \cos(x) \}}{dx} = - \sin(x) \\ \\ \sf \mapsto { \sin}^{2} (x) + { \cos}^{2} (x) = 1 \\ \\ \sf \mapsto \sin(2x) = 2 \sin(x) \cos(x)$