Question

Find the value of ∆

sin (ax + b).​

Answer 1

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Answer 2

Δ $= \dfrac{dy}{dx} =a\cos(ax+b)$

Step-by-step explanation:

We have,

$y=\sin(ax+b)$

To find, $\dfrac{dy}{dx}=?$

∴ $y=\sin(ax+b)$

Differentiating both sides w.r.t. x, we get

$\dfrac{dy}{dx} =\dfrac{d(\sin(ax+b))}{dx}$

⇒ $\dfrac{dy}{dx} =\cos(ax+b).\dfrac{d(ax+b)}{dx}$

⇒ $\dfrac{dy}{dx} =\cos(ax+b).(a+0)$

⇒ $\dfrac{dy}{dx} =a\cos(ax+b)$

∴ Δ $=\dfrac{dy}{dx} =a\cos(ax+b)$

Hence, $\dfrac{dy}{dx}=a\cos(ax+b)$