Mathematics, 17.09.2019 21:30 cutie22327
What is the derivative of 3x × sin^2 x^(1/2)
Answers: 1
![\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}\!\left[3x\cdot sin^2\big(x^{1/2}\big)\right]}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(3x)\cdot sin^2\big(x^{1/2}\big)+3x\cdot \dfrac{d}{dx}\!\left[sin^2\big(x^{1/2}\big)\right]}\\\\\\ \mathsf{\dfrac{dy}{dx}=3\cdot sin^2\big(x^{1/2}\big)+3x\cdot \dfrac{d}{dx}\!\left[sin^2\big(x^{1/2}\big)\right]}\\\\\\ \mathsf{\dfrac{dy}{dx}=3\cdot sin^2\big(x^{1/2}\big)+3x\cdot \dfrac{du}{dx}\qquad\quad(i)}](/tpl/images/0237/5338/689e7.png)
![\mathsf{\dfrac{du}{dx}=\dfrac{du}{dv}\cdot \dfrac{dv}{dw}\cdot \dfrac{dw}{dx}}\\\\\\ \mathsf{\dfrac{du}{dx}=\dfrac{d}{dv}(v^2)\cdot \dfrac{d}{dw}(sin\,w)\cdot \dfrac{d}{dx}\big(x^{1/2}\big)}\\\\\\ \mathsf{\dfrac{du}{dx}=\diagup\!\!\!\! 2v^{2-1}\cdot cos\,w\cdot \left[\dfrac{1}{\diagup\!\!\!\! 2}\,x^{(1/2)-1} \right]}\\\\\\ \mathsf{\dfrac{du}{dx}=v\cdot cos\,w\cdot x^{-1/2}}](/tpl/images/0237/5338/8c0e8.png)
![\mathsf{\dfrac{dy}{dx}=3\cdot sin^2\big(x^{1/2}\big)+3x\cdot \left[sin\big(x^{1/2}\big)\cdot cos\big(x^{1/2}\big)\cdot x^{-1/2}\right]}\\\\\\ \mathsf{\dfrac{dy}{dx}=3\cdot sin^2\big(x^{1/2}\big)+3x^{1-(1/2)}\cdot sin\big(x^{1/2}\big)\cdot cos\big(x^{1/2}\big)}](/tpl/images/0237/5338/6f01d.png)
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