Question

Y = cosθ, then dy/ dx at θ = π/2 will be

Answer 1

Given:

y= cosθ

To Find:

Differentiation of y w.r.t. θ or dy/dθ at θ= π/2

Solution:

On differentiating y w.r.t. θ, we get

$\longrightarrow\rm{\dfrac{dy}{d\theta}=-sin\theta}$

On putting θ= π/2 in dy/dx, we get

$\longrightarrow\rm{\bigg(\dfrac{dy}{d\theta}\bigg)_{\theta=\frac{\pi}{2}}=-sin\bigg(\dfrac{\pi}{2}\bigg)}$

$\longrightarrow\rm{\bigg(\dfrac{dy}{d\theta}\bigg)_{\theta=\frac{\pi}{2}}=-(1)}$

$\longrightarrow\rm\pink{\bigg(\dfrac{dy}{d\theta}\bigg)_{\theta=\frac{\pi}{2}}=-1}$

Hence, the value of dy/dθ at θ= π/2 is -1.

Formulae to Remember

$\longrightarrow\rm{\dfrac{d}{dx}(sinx)=cosx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosx)=-sinx}$

$\longrightarrow\rm{\dfrac{d}{dx}(tanx)=sec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(cotx)=-cosec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(secx)=secx.tanx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosecx)=-cosecx.cotx}$

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

$\longrightarrow\rm{\dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(e^{x})=e^{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(f(x))=f'(x)}$

Answer 2

Explanation:

since dcosФ/dx = -sinФ

also  y= cosФ

thus,  dy/dx = -sinФ= -sin(π/2) = -sin90° = -1  (∵Ф=π/2)