17 cm
Explanation explain
Answer:
28
Step-by-step explanation:
Multiply 35 by .8
Answer:
-15
Step-by-step explanation:
Answer:
11,875,000 common shares outstanding
Step-by-step explanation:
The number of common shares outstanding of Harper industries is determined by the company's common equity added to its MVA and then divided by the stock price per share.
If common equity is $900 million, MVA is $50 million and the price per share is $50, the number of shares outstanding is:
![n=\frac{900,000,000+50,000,000}{80}\\n=11,875,000](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B900%2C000%2C000%2B50%2C000%2C000%7D%7B80%7D%5C%5Cn%3D11%2C875%2C000)
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Find the derivative of
![\mathsf{y=3x\cdot sin^2\big(x^{1/2}\big)}](https://tex.z-dn.net/?f=%5Cmathsf%7By%3D3x%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%7D)
First, differentiate it by applying the product rule:
![\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)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%5C%21%5Cleft%5B3x%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%283x%29%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%2B3x%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%5C%21%5Cleft%5Bsin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D3%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%2B3x%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%5C%21%5Cleft%5Bsin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D3%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%2B3x%5Ccdot%20%5Cdfrac%7Bdu%7D%7Bdx%7D%5Cqquad%5Cquad%28i%29%7D)
where
![\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.](https://tex.z-dn.net/?f=%5Cmathsf%7Bu%3Dsin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%7D%5C%5C%5C%5C%5C%5C%5Cleft%5C%7B%5C%21%20%5Cbegin%7Barray%7D%7Bl%7D%20%5Cmathsf%7Bu%3Dv%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7Bv%3Dsin%5C%2Cw%7D%5C%5C%5C%5C%20%5Cmathsf%7Bw%3Dx%5E%7B1%2F2%7D%7D%20%5Cend%7Barray%7D%20%5Cright.)
As
u is a composite function, then you have to apply the chain rule to evaluate its derivative:
![\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}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdu%7D%7Bdx%7D%3D%5Cdfrac%7Bdu%7D%7Bdv%7D%5Ccdot%20%5Cdfrac%7Bdv%7D%7Bdw%7D%5Ccdot%20%5Cdfrac%7Bdw%7D%7Bdx%7D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bdu%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdv%7D%28v%5E2%29%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdw%7D%28sin%5C%2Cw%29%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bdu%7D%7Bdx%7D%3D%5Cdiagup%5C%21%5C%21%5C%21%5C%21%202v%5E%7B2-1%7D%5Ccdot%20cos%5C%2Cw%5Ccdot%20%5Cleft%5B%5Cdfrac%7B1%7D%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%202%7D%5C%2Cx%5E%7B%281%2F2%29-1%7D%20%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bdu%7D%7Bdx%7D%3Dv%5Ccdot%20cos%5C%2Cw%5Ccdot%20x%5E%7B-1%2F2%7D%7D)
Substitute back for
![\mathsf{v=sin\,w:}](https://tex.z-dn.net/?f=%5Cmathsf%7Bv%3Dsin%5C%2Cw%3A%7D)
![\mathsf{\dfrac{du}{dx}=sin\,w\cdot cos\,w\cdot x^{-1/2}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdu%7D%7Bdx%7D%3Dsin%5C%2Cw%5Ccdot%20cos%5C%2Cw%5Ccdot%20x%5E%7B-1%2F2%7D%7D)
and then substitute back for
![\mathsf{w=x^{1/2}:}](https://tex.z-dn.net/?f=%5Cmathsf%7Bw%3Dx%5E%7B1%2F2%7D%3A%7D)
![\mathsf{\dfrac{du}{dx}=sin\big(x^{1/2}\big)\cdot cos\big(x^{1/2}\big)\cdot x^{-1/2}\qquad\quad(ii)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdu%7D%7Bdx%7D%3Dsin%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Ccdot%20cos%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Ccdot%20x%5E%7B-1%2F2%7D%5Cqquad%5Cquad%28ii%29%7D)
Subsitute
(ii) into
(i) for
du/dx and you finally get
![\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)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D3%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%2B3x%5Ccdot%20%5Cleft%5Bsin%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Ccdot%20cos%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Ccdot%20x%5E%7B-1%2F2%7D%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D3%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%2B3x%5E%7B1-%281%2F2%29%7D%5Ccdot%20sin%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Ccdot%20cos%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%7D)
![\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.}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D3%5Ccdot%20sin%5E2%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%2B3x%5E%7B1%2F2%7D%5Ccdot%20sin%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%5Ccdot%20cos%5Cbig%28x%5E%7B1%2F2%7D%5Cbig%29%7D%5Cquad%5Clongleftarrow%5Cquad%5Ctextsf%7Bthis%20is%20the%20answer.%7D)
I hope this helps. =)
Tags: <em>derivative composite function product polynomial trigonometric trig sine sin power rule product rule chain rule differential integral calculus</em>