Question

if fx = sinx° find dy/dx

Answer 1

Appropriate Question:

$\sf \: If \: y \: = \: sin {x}^{ \circ}, \: find \: \dfrac{dy}{dx}$

Answer:

$\boxed{\sf \: \:\dfrac{dy}{dx} \: = \: \dfrac{\pi }{180}\:cos {x}^{ \circ} \: \: }$

Step-by-step explanation:

Given that,

$\sf \: \: y \: = \: sin {x}^{ \circ}$

can be rewritten as

$\sf \: \: y \: = \: sin \dfrac{\pi x}{180}$

On differentiating both sides w. r. t. x, we get

$\sf \: \:\dfrac{dy}{dx} \: = \:\dfrac{d}{dx}\left( sin \dfrac{\pi x}{180}\right)$

We know,

$\boxed{\sf \: \dfrac{d}{dx}sinx = cosx \: }$

So, using this result, we get

$\sf \: \:\dfrac{dy}{dx} \: = \:cos \dfrac{\pi x}{180}\dfrac{d}{dx}\left(\dfrac{\pi x}{180}\right)$

$\sf \: \:\dfrac{dy}{dx} \: = \:cos \dfrac{\pi x}{180} \times \dfrac{\pi }{180} \times \dfrac{d}{dx}(x)$

We know,

$\boxed{\sf \: \dfrac{d}{dx}x = 1 \: }$

So, using this result, we get

$\sf \: \:\dfrac{dy}{dx} \: = \:cos \dfrac{\pi x}{180} \times \dfrac{\pi }{180} \times 1$

$\implies\sf \: \sf \: \:\dfrac{dy}{dx} \: = \: \dfrac{\pi }{180}\:cos {x}^{ \circ}$

$\rule{190pt}{2pt}$

Additional Information

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {x}^{n} & \sf {nx}^{n - 1}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$