Question

please give right answer.its humble request 66​

Answer 1

GIVEN :–

$\\ \implies \bf y = \sin^{m} (ax)\cos^{m} (bx)$

TO FIND :–

$\\ \implies \bf \dfrac{dy}{dx} =?$

SOLUTION :–

$\\ \implies \bf y = \sin^{m} (ax)\cos^{m} (bx)$

• Differentiate with respect to 'x' –

$\\ \implies \bf \dfrac{dy}{dx}= \dfrac{d}{dx}[\sin^{m} (ax)\cos^{m} (bx)]$

• We know that –

$\\ \longrightarrow \red{\bf \dfrac{d}{dx}(u.v) = u \dfrac{dv}{dx} +v\dfrac{du}{dx}}$

• So that –

$\\ \implies \bf \dfrac{dy}{dx}= \sin^{m} (ax) \dfrac{d}{dx}[\cos^{m} (bx)] +\cos^{m} (bx) \dfrac{d}{dx}[\sin^{m} (ax)]$

$\\ \implies \bf \dfrac{dy}{dx}= \sin^{m} (ax) \left[m\cos^{m - 1} (bx) \times -b\sin(bx)\right] +\cos^{m} (bx)\left[m\sin^{m - 1} (ax) \times \cos(ax)\right]$

$\\ \implies \bf \dfrac{dy}{dx}=m\sin^{m - 1} (ax)\cos^{m - 1} (bx) \left[- b\sin(ax) \sin(bx)+a\cos(ax) \cos(bx)\right]$

$\\ \implies \bf \dfrac{dy}{dx}=m\sin^{m - 1} (ax)\cos^{m - 1} (bx) \left[-b\sin(ax) \sin(bx) +a\cos(ax) \cos(bx)\right]$

$\\ \implies \large{ \boxed{\bf \dfrac{dy}{dx}=m\sin^{m - 1} (ax)\cos^{m - 1} (bx) \left[a\cos(ax) \cos(bx)-b\sin(ax) \sin(bx) \right]}}$