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