# Gan's questions at Yahoo! Answers regarding differentiation

#### MarkFL

Staff member
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

Staff member
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.