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Gan's questions at Yahoo! Answers regarding differentiation
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Hello Gan,
1.) Differentiate the following with respect to $x$:
a) \(\displaystyle y^x\)
We could use the identity \(\displaystyle u=a^{\log_a(u)}\) to write:
\(\displaystyle y^x=e^{\ln\left(y^x \right)}\)
Next, we may use the logarithmic property \(\displaystyle \log_a\left(b^c \right)=c\cdot\log_a(b)\) to obtain:
\(\displaystyle y^x=e^{x\cdot\ln\left(y \right)}\)
Now we may differentiate, using the rule for the exponential function and the chain, product and logarithmic rules:
\(\displaystyle \frac{d}{dx}\left(y^x \right)=e^{x\cdot\ln\left(y \right)}\left(\frac{x}{y}+\ln(y) \right)\)
Since \(\displaystyle y^x=e^{x\cdot\ln\left(y \right)}\), we may now write:
\(\displaystyle \frac{d}{dx}\left(y^x \right)=y^x\left(\frac{x}{y}+\ln(y) \right)\)
Another method we could use is to write:
\(\displaystyle u=y^x\)
Take the natural log of both sides:
\(\displaystyle \ln(u)=\ln\left(y^x \right)=x\cdot\ln(y)\)
Implicitly differentiate with resepct to $x$:
\(\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{x}{y}+\ln(y)\)
Multiply through by $u$:
\(\displaystyle \frac{du}{dx}=u\left(\frac{x}{y}+\ln(y) \right)\)
Replace $u$ with $y^x$:
\(\displaystyle \frac{d}{dx}\left(y^x \right)=y^x\left(\frac{x}{y}+\ln(y) \right)\)
b) \(\displaystyle x^y=\sin(x)\)
Take the natural log of both sides:
\(\displaystyle y\ln(x)=\ln\left(\sin(x) \right)\)
Implicitly differentiate with respect to $x$:
\(\displaystyle \frac{y}{x}+\frac{dy}{dx}\ln(x)=\cot(x)\)
Solve for \(\displaystyle \frac{dy}{dx}\) to get:
\(\displaystyle \frac{dy}{dx}=\frac{x\cot(x)-y}{x\ln(x)}\)
2.) We are given $x^n+y^n=1$ and asked to find \(\displaystyle \frac{d^2y}{dx^2}\).
Implicitly differentiating the given equation with respect to $x$, we find:
\(\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx}=0\)
Solving for \(\displaystyle \frac{dy}{dx}\), we obtain:
\(\displaystyle \frac{dy}{dx}=-\frac{x^{n-1}}{y^{n-1}}\)
Differentiating again, we obtain:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{y^{n-1}\left((n-1)x^{n-2} \right)-x^{n-1}\left((n-1)y^{n-2}\dfrac{dy}{dx} \right)}{\left(y^{n-1} \right)^2}\)
Factoring the numerator and using \(\displaystyle \frac{dy}{dx}=-\frac{x^{n-1}}{y^{n-1}}\), we obtain:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{(n-1)x^{n-2}y^{n-2}\left(y-x\left(-\dfrac{x^{n-1}}{y^{n-1}} \right) \right)}{y^{2n-2}}\)
Combining terms in the rightmost factor in the numerator, we find:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{(n-1)x^{n-2}y^{n-2}\dfrac{x^{n}+y^n}{y^{n-1}}}{y^{2n-2}}\)
Using \(\displaystyle x^n+y^n=1\) and \(\displaystyle \frac{y^{n-2}}{y^{n-1}}=\frac{1}{y}\) we may write:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{(n-1)x^{n-2}}{y^{2n-1}}\)
Shown as desired.
1.) Differentiate the following with respect to $x$:
a) \(\displaystyle y^x\)
We could use the identity \(\displaystyle u=a^{\log_a(u)}\) to write:
\(\displaystyle y^x=e^{\ln\left(y^x \right)}\)
Next, we may use the logarithmic property \(\displaystyle \log_a\left(b^c \right)=c\cdot\log_a(b)\) to obtain:
\(\displaystyle y^x=e^{x\cdot\ln\left(y \right)}\)
Now we may differentiate, using the rule for the exponential function and the chain, product and logarithmic rules:
\(\displaystyle \frac{d}{dx}\left(y^x \right)=e^{x\cdot\ln\left(y \right)}\left(\frac{x}{y}+\ln(y) \right)\)
Since \(\displaystyle y^x=e^{x\cdot\ln\left(y \right)}\), we may now write:
\(\displaystyle \frac{d}{dx}\left(y^x \right)=y^x\left(\frac{x}{y}+\ln(y) \right)\)
Another method we could use is to write:
\(\displaystyle u=y^x\)
Take the natural log of both sides:
\(\displaystyle \ln(u)=\ln\left(y^x \right)=x\cdot\ln(y)\)
Implicitly differentiate with resepct to $x$:
\(\displaystyle \frac{1}{u}\frac{du}{dx}=\frac{x}{y}+\ln(y)\)
Multiply through by $u$:
\(\displaystyle \frac{du}{dx}=u\left(\frac{x}{y}+\ln(y) \right)\)
Replace $u$ with $y^x$:
\(\displaystyle \frac{d}{dx}\left(y^x \right)=y^x\left(\frac{x}{y}+\ln(y) \right)\)
b) \(\displaystyle x^y=\sin(x)\)
Take the natural log of both sides:
\(\displaystyle y\ln(x)=\ln\left(\sin(x) \right)\)
Implicitly differentiate with respect to $x$:
\(\displaystyle \frac{y}{x}+\frac{dy}{dx}\ln(x)=\cot(x)\)
Solve for \(\displaystyle \frac{dy}{dx}\) to get:
\(\displaystyle \frac{dy}{dx}=\frac{x\cot(x)-y}{x\ln(x)}\)
2.) We are given $x^n+y^n=1$ and asked to find \(\displaystyle \frac{d^2y}{dx^2}\).
Implicitly differentiating the given equation with respect to $x$, we find:
\(\displaystyle nx^{n-1}+ny^{n-1}\frac{dy}{dx}=0\)
Solving for \(\displaystyle \frac{dy}{dx}\), we obtain:
\(\displaystyle \frac{dy}{dx}=-\frac{x^{n-1}}{y^{n-1}}\)
Differentiating again, we obtain:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{y^{n-1}\left((n-1)x^{n-2} \right)-x^{n-1}\left((n-1)y^{n-2}\dfrac{dy}{dx} \right)}{\left(y^{n-1} \right)^2}\)
Factoring the numerator and using \(\displaystyle \frac{dy}{dx}=-\frac{x^{n-1}}{y^{n-1}}\), we obtain:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{(n-1)x^{n-2}y^{n-2}\left(y-x\left(-\dfrac{x^{n-1}}{y^{n-1}} \right) \right)}{y^{2n-2}}\)
Combining terms in the rightmost factor in the numerator, we find:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{(n-1)x^{n-2}y^{n-2}\dfrac{x^{n}+y^n}{y^{n-1}}}{y^{2n-2}}\)
Using \(\displaystyle x^n+y^n=1\) and \(\displaystyle \frac{y^{n-2}}{y^{n-1}}=\frac{1}{y}\) we may write:
\(\displaystyle \frac{d^2y}{dx^2}=-\frac{(n-1)x^{n-2}}{y^{2n-1}}\)
Shown as desired.