Hello Robert,
For the first problem, we are given the following curve along which a point moves:
$$x^2+y^2=144$$
We are asked to find $$\frac{dx}{dt}$$ at a particular point, so implicitly differentiating with respect to $t$, we find:
$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$
Solving for $$\frac{dx}{dt}$$, we obtain:
$$\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$$
Plugging in the given data:
$$x=-2\text{ cm},\,y=-\sqrt{140}\text{ cm}=-2\sqrt{35}\text{ cm},\,\frac{dy}{dt}=2\,\frac{\text{cm}}{\text{min}}$$
we have:
$$\frac{dx}{dt}=-\frac{-2\sqrt{35}\text{ cm}}{-2\text{ cm}}\cdot2\,\frac{\text{cm}}{\text{min}}=-2\sqrt{35}\,\frac{\text{cm}}{\text{min}}$$
For the second problem, we are given:
$$y=x^{x^6}+2$$
Differentiating with respect to $x$, we may write:
$$\frac{dy}{dx}=\frac{d}{dx}\left(x^{x^6} \right)+\frac{d}{dx}(2)$$
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:
$$\frac{dy}{dx}=\frac{d}{dx}\left(e^{x^6\ln(x)} \right)$$
Using the differentiation rule for the exponential function, the chain and product rules, we may write:
$$\frac{dy}{dx}=e^{x^6\ln(x)}\left(x^6\frac{1}{x}+6x^5\ln(x) \right)$$
Hence:
$$\frac{dy}{dx}=x^{x^6}\left(x^5+6x^5\ln(x) \right)$$
Factoring and using the rule for exponents $a^ba^c=a^{b+c}$:
$$\frac{dy}{dx}=x^{x^6+5}\left(1+6\ln(x) \right)$$