Question

Q1.Find
dy/dx if y= tan 5x

​

Answer 1

Answer:

5sec5x

Step-by-step explanation:

derivative of tan5x will be first

sec5x * d(5x)/dx

= 5sec5x

Mark it as the brainliest dear...

Answer 2

Step-by-step explanation:

Answer :

The derivative of

$\sf{tan(5x)}tan(5x) is \sf{5sec^{2}(5x)}$

Explanation :

Given :

Function of y =

$\sf{tan(5x)}$

To find :

The derivative of

$\sf{tan(5x)}$

$\sf{\dfrac{dy}{dx} = ?}$

Knowledge required :

Chain rule of differentiation :

$\boxed{\sf{\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}}}$

Differentiation of

$\sf{tan(x)}tan(x) is \sf{sec^{2}(x)}$

i.e,

$\boxed{\sf{\dfrac{d[tan(x)]}{dx} = sec^{2}(x)}}$

$\sf{tan(x) = \dfrac{sin(x)}{cos(x)}}$

$\sf{cos(x) = \dfrac{1}{sec(x)}}$

$\sf{sin^{2}(x) + cos^{2}(x) = 1}$

Solution :

By applying the chain rule of differentiation and substituting the values in it, we get :

(Here , taking as u as 5x)

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{d[tan5(x)]}{d(5x)} \times \dfrac{d(5x)}{dx}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = sec^{2}(5x) \times 5} \\ \\ :\implies \sf{\dfrac{dy}{dx} = 5sec^{2}(5x)} \\ \\ \boxed{\therefore \sf{\dfrac{d[tan(5x)]}{dx} = 5sec^{2}(5x)}} \\ \\\end{gathered}$

Therefore,

The derivative of

$\sf{tan(5x)}tan(5x) , \sf{\dfrac{dy}{dx} = 5sec^{2}(5x)}$

Alternative method :

By applying the quotient rule and substituting the values in it, We get :

$\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{v\dfrac{d(u)}{dx} - u\dfrac{d(v)}{dx}}{v^{2}}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{[cos(5x)]\dfrac{d[sin(5x)]}{dx} - [sin(5x)]\dfrac{d[cos(5x)]}{dx}}{cos^{2}(5x)}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{[cos(5x)] \times cos(5x) \times 5 - [sin(5x)] \times sin(5x) \times 5}{cos^{2}(5x)}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{5cos^{2}(5x) - (-)5sin^{2}(5x)}{cos^{2}(5x)}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{5cos^{2}(5x) + 5sin^{2}(5x)}{cos^{2}(5x)}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{5[cos^{2}(5x) + sin^{2}(5x)]}{cos^{2}(5x)}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{5(1)}{cos^{2}(5x)}} \\ \\ :\implies \sf{\dfrac{dy}{dx} = 5sec^{2}(5x)} \\ \\ \boxed{\therefore \sf{\dfrac{d[tan(5x)]}{dx} = 5sec^{2}(5x)}} \\ \\\end{gathered}$

Therefore,

The derivative of

$\sf{tan(5x)}tan(5x) ,\sf{\dfrac{dy}{dx} = 5sec^{2}(5x)}$