Question

find dy/dx if y=sec2x​

Answer 1

Knowledge required :

• Chain rule of differentiation :

$\boxed{\sf{\dfrac{dy}{dx} = \dfrac{d(y)}{d(u)} \times \dfrac{d(u)}{d(x)}}}$

• Exponential rule of differentiation :

$\boxed{\sf{\dfrac{d(x^{n})}{dx} = nx^{(n - 1)}}}$

Solution :

Given ,

• $\sf{f(y) = sec^{2}(x)}$

$\boxed{\sf{\dfrac{dy}{dx} = \dfrac{d(y)}{d(u)} \times \dfrac{d(u)}{d(x)}}}$

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

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(y)}{d(u)} \times \dfrac{d(u)}{d(x)}}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(sec^{2}(x))}{d(sec(x))} \times \dfrac{d(sec(x))}{dx}}$

Now by differentiating sec²x , we get :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(sec^{2}(x))}{d(sec x)}}$

By applying the rule of , x^n = nx^(n - 1) , we get :

$:\implies \sf{\dfrac{dy}{dx} = 2 \times sec^{(2 - 1)}(x)}$

$:\implies \sf{\dfrac{dy}{dx} = 2sec^{1}(x)}$

$:\implies \sf{\dfrac{dy}{dx} = 2 sec(x)}$

$\boxed{\therefore \sf{\dfrac{d(sec^{2}(x))}{d(sec (x))} = 2 sec(x)}}$

By differentiating sec x , we get :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(sec(x))}{dx}}$

We know that the differentiation of sec x i.e, sec(x)tan(x).

By substituting it in the equation , we get :

$:\implies \sf{\dfrac{dy}{dx} = sec(x)tan(x)}$

$\boxed{\therefore \sf{\dfrac{d(sec(x))}{dx} = sec(x)tan(x)}}$

Now we know the derivative of sec²x and sec x , so by substituting it in the equation , :

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d(sec^{2}(x))}{d(sec(x))} \times \dfrac{d(sec(x))}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = 2sec(x) \times sec(x)tan(x)}$

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

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

Hence the derivative of sec²x is 2 sec²xtanx.