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jeyben [28]
3 years ago
9

Mary invests £12,000 in a savings account the account pays 1.5% compound interest per year work out the value of her investment

after 2 years.​
Mathematics
1 answer:
Lesechka [4]3 years ago
5 0

Answer:

The value of Mary's investment after two years = £12362.7

Step-by-step explanation:

P = Principal / initial amount

R = rate of interest per cent per year

T = number of years

A = final amount at the end of T years

Then:

A = P*(1 + R/100)^2

In our example:

P = £12000

R = 1.5 per cent per year

T = 2 years

Thus:

A = 12000*(1 + 1.5/100)^2

  = 12000*(1 + 0.015)^2

  = 12000*(1.015)^2

   = 12000*(1.030225)

   = 12362.7

Value of investment after two years = £12362.7

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Find thd <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D" id="TexFormula1" title="\frac{dy}{dx}" alt="\frac{dy}{dx}" a
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x^3y^2+\sin(x\ln y)+e^{xy}=0

Differentiate both sides, treating y as a function of x. Let's take it one term at a time.

Power, product and chain rules:

\dfrac{\mathrm d(x^3y^2)}{\mathrm dx}=\dfrac{\mathrm d(x^3)}{\mathrm dx}y^2+x^3\dfrac{\mathrm d(y^2)}{\mathrm dx}

=3x^2y^2+x^3(2y)\dfrac{\mathrm dy}{\mathrm dx}

=3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(\sin(x\ln y)}{\mathrm dx}=\cos(x\ln y)\dfrac{\mathrm d(x\ln y)}{\mathrm dx}

=\cos(x\ln y)\left(\dfrac{\mathrm d(x)}{\mathrm dx}\ln y+x\dfrac{\mathrm d(\ln y)}{\mathrm dx}\right)

=\cos(x\ln y)\left(\ln y+\dfrac1y\dfrac{\mathrm dy}{\mathrm dx}\right)

=\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}

Product and chain rules:

\dfrac{\mathrm d(e^{xy})}{\mathrm dx}=e^{xy}\dfrac{\mathrm d(xy)}{\mathrm dx}

=e^{xy}\left(\dfrac{\mathrm d(x)}{\mathrm dx}y+x\dfrac{\mathrm d(y)}{\mathrm dx}\right)

=e^{xy}\left(y+x\dfrac{\mathrm dy}{\mathrm dx}\right)

=ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}

The derivative of 0 is, of course, 0. So we have, upon differentiating everything,

3x^2y^2+6x^3y\dfrac{\mathrm dy}{\mathrm dx}+\cos(x\ln y)\ln y+\dfrac{\cos(x\ln y)}y\dfrac{\mathrm dy}{\mathrm dx}+ye^{xy}+xe^{xy}\dfrac{\mathrm dy}{\mathrm dx}=0

Isolate the derivative, and solve for it:

\left(6x^3y+\dfrac{\cos(x\ln y)}y+xe^{xy}\right)\dfrac{\mathrm dy}{\mathrm dx}=-\left(3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}\right)

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^2+\cos(x\ln y)\ln y-ye^{xy}}{6x^3y+\frac{\cos(x\ln y)}y+xe^{xy}}

(See comment below; all the 6s should be 2s)

We can simplify this a bit by multiplying the numerator and denominator by y to get rid of that fraction in the denominator.

\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{3x^2y^3+y\cos(x\ln y)\ln y-y^2e^{xy}}{6x^3y^2+\cos(x\ln y)+xye^{xy}}

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3 years ago
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