Take the third equation and divide it by 2. If the equations are the same, they have infinitely many solutions because every point on one eq. is the same on the other. You can say they overlap.
2% interest is the rate "r" but needs to be decimal form. r = 0.02
$400 is the principal amount P
because it's annual rate, n = 1
t = 6
A = 400(1+0.02)^(6)
A = 400(1.02)^(6)
A = 400(1.13)
A = 450.46
Answer:
the ratio is 18:36 = 1:2. Simply reduced
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You can actually use either the product rule or the chain rule for this one. Observe:
• Method I:y = cos² xy = cos x · cos xDifferentiate it by applying the product rule:
![\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%5Ccdot%20cos%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%5Ccdot%20cos%5C%2Cx%2Bcos%5C%2Cx%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%7D)
The derivative of
cos x is
– sin x. So you have
![\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%28-sin%5C%2Cx%29%5Ccdot%20cos%5C%2Cx%2Bcos%5C%2Cx%5Ccdot%20%28-sin%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D-sin%5C%2Cx%5Ccdot%20cos%5C%2Cx-cos%5C%2Cx%5Ccdot%20sin%5C%2Cx%7D)
![\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Ctherefore~~%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D-2%5C%2Csin%5C%2Cx%5Ccdot%20cos%5C%2Cx%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
—————
• Method II:You can also treat
y as a composite function:
![\left\{\! \begin{array}{l} \mathsf{y=u^2}\\\\ \mathsf{u=cos\,x} \end{array} \right.](https://tex.z-dn.net/?f=%5Cleft%5C%7B%5C%21%0A%5Cbegin%7Barray%7D%7Bl%7D%0A%5Cmathsf%7By%3Du%5E2%7D%5C%5C%5C%5C%0A%5Cmathsf%7Bu%3Dcos%5C%2Cx%7D%0A%5Cend%7Barray%7D%0A%5Cright.)
and then, differentiate
y by applying the chain rule:
![\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bdy%7D%7Bdu%7D%5Ccdot%20%5Cdfrac%7Bdu%7D%7Bdx%7D%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D%5Cdfrac%7Bd%7D%7Bdu%7D%28u%5E2%29%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%7D)
For that first derivative with respect to
u, just use the power rule, then you have
![\mathsf{\dfrac{dy}{dx}=2u^{2-1}\cdot \dfrac{d}{dx}(cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=2u\cdot (-sin\,x)\qquad\quad (but~~u=cos\,x)}\\\\\\ \mathsf{\dfrac{dy}{dx}=2\,cos\,x\cdot (-sin\,x)}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D2u%5E%7B2-1%7D%5Ccdot%20%5Cdfrac%7Bd%7D%7Bdx%7D%28cos%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D2u%5Ccdot%20%28-sin%5C%2Cx%29%5Cqquad%5Cquad%20%28but~~u%3Dcos%5C%2Cx%29%7D%5C%5C%5C%5C%5C%5C%0A%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D2%5C%2Ccos%5C%2Cx%5Ccdot%20%28-sin%5C%2Cx%29%7D)
and then you get the same answer:
![\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Ctherefore~~%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7B%5Cdfrac%7Bdy%7D%7Bdx%7D%3D-2%5C%2Csin%5C%2Cx%5Ccdot%20cos%5C%2Cx%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
I hope this helps. =)
Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>