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Robert's questions at Yahoo! Answers regarding differentiation

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MarkFL

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Feb 24, 2012
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Here are the questions:

Please help solve these calculus questions quick tmrw?


A point moves along the circle x^2+y^2=144so that dy/dt=2cm/min. Find dx/dt at the point (-2,−√140).

Given y=x^x^6+2, find dy/dx
I have posted a link there to this topic so the OP can see my work.
 
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MarkFL

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Feb 24, 2012
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Hello Robert,

For the first problem, we are given the following curve along which a point moves:

\(\displaystyle x^2+y^2=144\)

We are asked to find \(\displaystyle \frac{dx}{dt}\) at a particular point, so implicitly differentiating with respect to $t$, we find:

\(\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt}=0\)

Solving for \(\displaystyle \frac{dx}{dt}\), we obtain:

\(\displaystyle \frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}\)

Plugging in the given data:

\(\displaystyle x=-2\text{ cm},\,y=-\sqrt{140}\text{ cm}=-2\sqrt{35}\text{ cm},\,\frac{dy}{dt}=2\,\frac{\text{cm}}{\text{min}}\)

we have:

\(\displaystyle \frac{dx}{dt}=-\frac{-2\sqrt{35}\text{ cm}}{-2\text{ cm}}\cdot2\,\frac{\text{cm}}{\text{min}}=-2\sqrt{35}\,\frac{\text{cm}}{\text{min}}\)

For the second problem, we are given:

\(\displaystyle y=x^{x^6}+2\)

Differentiating with respect to $x$, we may write:

\(\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(x^{x^6} \right)+\frac{d}{dx}(2)\)

Let's rewrite the first term on the right using an exponential/logarithmic identity and observing that for the second term, the derivative of a constant is zero, hence:

\(\displaystyle \frac{dy}{dx}=\frac{d}{dx}\left(e^{x^6\ln(x)} \right)\)

Using the differentiation rule for the exponential function, the chain and product rules, we may write:

\(\displaystyle \frac{dy}{dx}=e^{x^6\ln(x)}\left(x^6\frac{1}{x}+6x^5\ln(x) \right)\)

Hence:

\(\displaystyle \frac{dy}{dx}=x^{x^6}\left(x^5+6x^5\ln(x) \right)\)

Factoring and using the rule for exponents $a^ba^c=a^{b+c}$:

\(\displaystyle \frac{dy}{dx}=x^{x^6+5}\left(1+6\ln(x) \right)\)