Question

secx + tan y =1 than find dy/dx at origin
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Answer 1

Given :

• secx + tan y =1

To find :

$\sf \: \dfrac{dy}{dx} \: \: at \: \: origin$

Solution :

$\implies \sf \: sec \: x + tan \: y =1$

Now , differentiating both sides :-

$\implies \sf \: \dfrac{d(sec \: x + tan \: y)}{dx} = \dfrac{d(1)}{dx}$

we know that :-

$\underline{\boxed{\bf\dfrac{d(f(x) + g( y))}{dx} = \dfrac{d(f(x) )}{dx} + \dfrac{d( g( y))}{dx} }}$

$\underline{\boxed{\bf\dfrac{d(c)}{dx} = 0 }} \\ \sf where \: \bf c \sf\: is \: constant \: term$

$\implies \sf \: \dfrac{d(sec \: x )}{dx} + \dfrac{d( tan \: y)}{dx} =0$

We know that :-

$\underline{\boxed{\bf\dfrac{d(sec \: x)}{dx} =sec \: x \times tan \: x }}$

$\underline{\boxed{\bf\dfrac{d(tan\: y)}{dx} = {sec \: }^{2} y \: \dfrac{dy}{dx} }}$

$\implies\sf (sec \: x\times tan \: x)+( {sec }^{2}y)\dfrac{dy}{dx} =0$

Now , coordinates of origin = (0,0)

Therefore put x and y = 0°

$\implies\sf (sec 0 \degree \times tan 0\degree)+( {sec }^{2}0\degree)\dfrac{dy}{dx} =0$

We know that :-

$\underline{\boxed{ \bf sec \: 0 \degree = 1}}$

$\underline{\boxed{ \bf tan \: 0 \degree = 0}}$

$\implies\sf (1 \times 0)+ {( 1)}^{2} \dfrac{dy}{dx} =0$

$\implies\sf 0+ {( 1)}^{2} \dfrac{dy}{dx} =0$

$\implies\sf {( 1)} \times \dfrac{dy}{dx} =0$

$\implies\sf \dfrac{dy}{dx} =0$

Answer : 0