Question

if y=tan^x find dy/dx
​

Answer 1

Step-by-step explanation:

Given:

y = tanx

TO Find: The value of dy/dx

Formula Used:

$\tan(x) = \frac{ \sin(x) }{ \cos(x) }$

$\frac{d}{dx} ( \frac{u}{v} ) = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {v}^{2} }$

Solution:

$\frac{d}{dx} \tan(x) \\ = \frac{d}{dx} \frac{ \sin(x) }{ \cos(x) } \\ = \frac{ \cos(x) \frac{d}{dx} \sin(x) - \sin(x) \frac{d}{dx} \cos(x) }{ { \cos}^{2} x} \\ = \frac{ \cos(x) \cos(x) - \sin(x)( - \sin(x) )}{ { \cos}^{2}x } \\ = \frac{ { \cos}^{2} x + { \sin}^{2}x }{ { \cos }^{2}x } \\ = \frac{1}{ { \cos }^{2} x } \\ = { \sec }^{2} x$

$\huge\mathbb{\purple{\underline{\red{\underline{\pink{✴answer{\orange{ \: \: :{\blue{ { \sec}^{2} x}}}}}}}}}}</p><p> \\ \:$

Answer 2

Answer:

Given:

y = tanx

TO Find: The value of dy/dx

Formula Used:

\tan(x) = \frac{ \sin(x) }{ \cos(x) }tan(x)=

cos(x)

sin(x)

\frac{d}{dx} ( \frac{u}{v} ) = \frac{v \frac{du}{dx} - u \frac{dv}{dx} }{ {v}^{2} }

dx

d

(

v

u

)=

v

2

v

dx

du

−u

dx

dv

Solution:

\begin{gathered} \frac{d}{dx} \tan(x) \\ = \frac{d}{dx} \frac{ \sin(x) }{ \cos(x) } \\ = \frac{ \cos(x) \frac{d}{dx} \sin(x) - \sin(x) \frac{d}{dx} \cos(x) }{ { \cos}^{2} x} \\ = \frac{ \cos(x) \cos(x) - \sin(x)( - \sin(x) )}{ { \cos}^{2}x } \\ = \frac{ { \cos}^{2} x + { \sin}^{2}x }{ { \cos }^{2}x } \\ = \frac{1}{ { \cos }^{2} x } \\ = { \sec }^{2} x\end{gathered}

dx

d

tan(x)

=

dx

d

cos(x)

sin(x)

=

cos

2

x

cos(x)

dx

d

sin(x)−sin(x)

dx

d

cos(x)

=

cos

2

x

cos(x)cos(x)−sin(x)(−sin(x))

=

cos

2

x

cos

2

x+sin

2

x

=

cos

2

x

1

=sec

2

x

okay hope it's helpful to you dear