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Find the derivative of
[tex]\mathsf{y=3x\cdot sin^2\big(x^{1/2}\big)}[/tex]
First, differentiate it by applying the product rule:
[tex]\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)}[/tex]
where
[tex]\mathsf{u=sin^2\big(x^{1/2}\big)}\\\\\\\left\{\! \begin{array}{l} \mathsf{u=v^2}\\\\ \mathsf{v=sin\,w}\\\\ \mathsf{w=x^{1/2}} \end{array} \right.[/tex]
As u is a composite function, then you have to apply the chain rule to evaluate its derivative:
[tex]\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}}[/tex]
Substitute back for [tex]\mathsf{v=sin\,w:}[/tex]
[tex]\mathsf{\dfrac{du}{dx}=sin\,w\cdot cos\,w\cdot x^{-1/2}}[/tex]
and then substitute back for [tex]\mathsf{w=x^{1/2}:}[/tex]
[tex]\mathsf{\dfrac{du}{dx}=sin\big(x^{1/2}\big)\cdot cos\big(x^{1/2}\big)\cdot x^{-1/2}\qquad\quad(ii)}[/tex]
Subsitute (ii) into (i) for du/dx and you finally get
[tex]\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)}[/tex]
[tex]\mathsf{\dfrac{dy}{dx}=3\cdot sin^2\big(x^{1/2}\big)+3x^{1/2}\cdot sin\big(x^{1/2}\big)\cdot cos\big(x^{1/2}\big)}\quad\longleftarrow\quad\textsf{this is the answer.}[/tex]
I hope this helps. =)
Tags: derivative composite function product polynomial trigonometric trig sine sin power rule product rule chain rule differential integral calculus
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Find the derivative of
[tex]\mathsf{y=3x\cdot sin^2\big(x^{1/2}\big)}[/tex]
First, differentiate it by applying the product rule:
[tex]\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)}[/tex]
where
[tex]\mathsf{u=sin^2\big(x^{1/2}\big)}\\\\\\\left\{\! \begin{array}{l} \mathsf{u=v^2}\\\\ \mathsf{v=sin\,w}\\\\ \mathsf{w=x^{1/2}} \end{array} \right.[/tex]
As u is a composite function, then you have to apply the chain rule to evaluate its derivative:
[tex]\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}}[/tex]
Substitute back for [tex]\mathsf{v=sin\,w:}[/tex]
[tex]\mathsf{\dfrac{du}{dx}=sin\,w\cdot cos\,w\cdot x^{-1/2}}[/tex]
and then substitute back for [tex]\mathsf{w=x^{1/2}:}[/tex]
[tex]\mathsf{\dfrac{du}{dx}=sin\big(x^{1/2}\big)\cdot cos\big(x^{1/2}\big)\cdot x^{-1/2}\qquad\quad(ii)}[/tex]
Subsitute (ii) into (i) for du/dx and you finally get
[tex]\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)}[/tex]
[tex]\mathsf{\dfrac{dy}{dx}=3\cdot sin^2\big(x^{1/2}\big)+3x^{1/2}\cdot sin\big(x^{1/2}\big)\cdot cos\big(x^{1/2}\big)}\quad\longleftarrow\quad\textsf{this is the answer.}[/tex]
I hope this helps. =)
Tags: derivative composite function product polynomial trigonometric trig sine sin power rule product rule chain rule differential integral calculus