Gan's questions at Yahoo Answers regarding differentiation

[MarkFL]

Here are the questions:

Differentiate the following equation with respect to x. (a) y^x (b)x^y= sin x?


Differentiate the following equation with respect to x. (a) y^x (b)x^y= sin x?
given that x^n + y^n = 1, show that d2y/dx2 = -(n-1)x^(n-2)/ y^(2n-1)

I have posted a link there to this thread so the OP can see my work.

 

[MarkFL]

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.