MHB Robert's questions at Yahoo Answers regarding differentiation

[MarkFL]

Here are the questions:

Please help solve these calculus questions quick tmrw?


A point moves along the circle x^2+y^2=144so that dy/dt=2cm/min. Find dx/dt at the point (-2,−√140).

Given y=x^x^6+2, find dy/dx

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

 

[MarkFL]

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 $a^ba^c=a^{b+c}$:

$$\frac{dy}{dx}=x^{x^6+5}\left(1+6\ln(x) \right)$$