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Robert's questions at Yahoo! Answers regarding differentiation
- Thread starter MarkFL
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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 $a^ba^c=a^{b+c}$:
\(\displaystyle \frac{dy}{dx}=x^{x^6+5}\left(1+6\ln(x) \right)\)
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 $a^ba^c=a^{b+c}$:
\(\displaystyle \frac{dy}{dx}=x^{x^6+5}\left(1+6\ln(x) \right)\)