Question

differentiate w r t x if y=cos (x^2 + a^2)​

Answer 1

Given :

y=cos(x²+ a²)

To find

dy /dx

Theory :

Chain rule

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\sf\:\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Solution :

We have ,

$\sf\:y=\cos(x^2+a^2)$

$\sf\dfrac{dy}{dx}=\dfrac{d\cos(x^2+a^2)}{d(x^2+a^2)}\times\dfrac{x^2+a^2}{dx}$

$\sf\dfrac{dy}{dx}=\dfrac{d\cos(x^2+a^2)}{d(x^2+a^2)}\times[\dfrac{x^2}{dx}+\dfrac{a^2}{dx}]$

$\sf\dfrac{dy}{dx}=-\sin(x^2+a^2)\times[2x+0]$

$\sf\dfrac{dy}{dx}=-\sin(x^2+a^2)\times2x$

It is required solution!

$\rule{200}2$

${\underline{\sf{Formula's}}}$

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

$2)\sf\:\frac{d(constant)}{dx}=0$

$3)\sf\dfrac{d(\cos)}{dx}=-sin\:x$

Answer 2

Given ,

The function is y = Cos(x² + a²)

Differentiating with respect to x , we get

$\sf \mapsto \frac{dy}{dx} = \frac{dCos( {x}^{2} + {a}^{2} )}{dy} \\ \\ \sf \mapsto \frac{dy}{dx} = -Sin( {x}^{2} + {a}^{2}) \times ( \frac{d( {x}^{2} + {a}^{2}) }{dx} ) \\ \\ \sf \mapsto \frac{dy}{dx} = -Sin( {x}^{2} + {a}^{2}) \times 2x$

Remmember :

$\sf \mapsto \frac{d \{Cos(x) \}}{dx} = - Sin(x) \\ \\ \sf \mapsto \frac{d {(x)}^{n} }{dx} = n {(x)}^{n - 1}\\ \\ \sf \mapsto \frac{d (Constant)}{dx} = 0$