We start by taking the derivative of x and y with respect to t, as both of the equations are only in terms of this variable:

\(\displaystyle \frac{dx}{dt}=6t^2+10t-11\)

\(\displaystyle \frac{dy}{dt}=14t-4\)

The problem asks us to find the derivative of the parametric equations, dy/dx, and we can see from the work below that the dt term is cancelled when we divide dy/dt by dx/dt, leaving us with dy/dx:

\(\displaystyle \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dt}\frac{dt}{dx}=\frac{dy}{dx}\)

So now that we know dx/dt and dy/dt, all we must do to find the derivative of our parametric equations is divide dy/dt by dx/dt:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{14t-4}{6t^2+10t-11}=\frac{2(7t-2)}{6t^2+10t-11}\)

We can determine that \(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)  since the \(\displaystyle dt\) terms will cancel out in the division process.

Since \(\displaystyle x=2t-5\) and \(\displaystyle y=3t+8\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to derive

\(\displaystyle \frac{dx}{dt}=(1)2t^{1-1}-(0)5=2\)and \(\displaystyle \frac{dy}{dt}=(1)3t^{1-1}+(0)8=3\) .

Thus:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3}{2}}\).

Our first step in finding the derivative dy/dx of the polar equation is to find the derivative of r with respect to \(\displaystyle \theta\). This gives us:

\(\displaystyle \frac{dr}{d\theta}=8cos\theta\)

Now that we know dr/d\(\displaystyle \theta\), we can plug this value into the equation for the derivative of an expression in polar form:

\(\displaystyle \frac{dy}{dx}=\frac{\frac{dr}{d\theta}sin\theta+rcos\theta }{\frac{dr}{d\theta }cos\theta-rsin\theta }=\frac{8cos\theta sin\theta+(3+8sin\theta )cos\theta}{8cos^2\theta-(3+8sin\theta)sin\theta }\)

Simplifying the equation, we get our final answer for the derivative of r:

\(\displaystyle \frac{dy}{dx}=\frac{16cos\theta sin\theta +3cos\theta }{8cos^2\theta -3sin\theta -8sin^2\theta }\)

The derivative of a polar function is found using the formula

\(\displaystyle \frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}\)

The only unknown piece is \(\displaystyle r'(\theta)\). Recall that the derivative of a constant is zero, and that 

\(\displaystyle \frac{d}{d\theta}sin(\theta)=cos(\theta)\), so

\(\displaystyle r'(\theta)=-4cos(\theta)\)

Substiting \(\displaystyle r'(\theta)\ and\ r(\theta)\) this into the derivative formula, we find

\(\displaystyle \frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos(\theta)cos(\theta)-(3-4sin(\theta))sin(\theta)}\)

\(\displaystyle \frac{dy}{dx}=\frac{-4cos(\theta)sin(\theta)+(3-4sin(\theta))cos(\theta)}{-4cos^2(\theta)-(3-4sin(\theta))sin(\theta)}\)

In general, the dervative of a function in polar coordinates can be written as

\(\displaystyle \frac{dy}{dx}=\frac{r'(\theta)sin(\theta)+r(\theta)cos(\theta)}{r'(\theta)cos(\theta)-r(\theta)sin(\theta)}\).

Therefore, we need to find \(\displaystyle r'(\theta)\), and then substitute \(\displaystyle r(\theta)\ and\ r'(\theta)\) into the derivative formula.

To find \(\displaystyle r'(\theta)\), the chain rule, 

\(\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))g'(x)\), is necessary.

We also need to know that 

\(\displaystyle \frac{d}{dx}sin(x)=cos(x)\).

Therefore,

\(\displaystyle r'(\theta)=3cos(3\theta)\).  

Substituting \(\displaystyle r(\theta)\ and\ r'(\theta)\) into the derivative formula yields

\(\displaystyle \frac{dy}{dx}=\frac{3cos(3\theta)sin(\theta)+sin(3\theta)cos(\theta)}{3cos(3\theta)cos(\theta)-sin(3\theta)sin(\theta)}\)

In order to find the derivative   of a polar equation \(\displaystyle r=7sin\theta-6\), we must first find the derivative of with respect to  as follows:

\(\displaystyle \frac{dr}{d\theta}=7cos\theta\)

We can then swap the given values of  and  into the equation of the derivative of an expression into polar form:

\(\displaystyle \frac{dy}{dx}=\frac{7cos\theta\sin\theta+(7sin\theta-6)\cos\theta}{7cos\theta\cos\theta-(7sin\theta-6)\sin\theta}\)

\(\displaystyle \frac{dy}{dx}=\frac{7cos\theta\sin\theta+7sin\theta\cos\theta-6\\cos\theta}{7cos\theta\cos\theta-7sin\theta\sin\theta+6\sin\theta}\)

\(\displaystyle \frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7(cos^{2}\theta-sin^{2}\theta)+6\sin\theta}\)

Using the trigonometric identity , we can deduce that . Swapping this into the denominator, we get:

\(\displaystyle \frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7((1-\sin^{2}\theta)-sin^{2}\theta)+6\sin\theta}\)

\(\displaystyle \frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7(1-2\sin^{2}\theta)+6\sin\theta}\)

\(\displaystyle \frac{dy}{dx}=\frac{14cos\theta\sin\theta-6\cos\theta}{7-14\sin^{2}\theta+6\sin\theta}\)

By definition, to find the gradient vector , we will have to find the gradient components. We know that the gradient components are that partial derivatives.

We know that in our case we have :

\(\displaystyle \frac{\partial f}{\partial x_{i}}=\frac{x_{i}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\)

To see this, fix all other variables and assume that you have only \(\displaystyle x_{i}\) as the only variable.

Now we apply the given defintion , i.e,

\(\displaystyle \nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]\)

with :

\(\displaystyle \frac{\partial f}{\partial x_{1}}=\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\)

\(\displaystyle \frac{\partial f}{\partial x_{2}}=\frac{x_{2}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\)

\(\displaystyle \frac{\partial f}{\partial x_{3}}=\frac{x_{3}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\)

\(\displaystyle \vdots\)

\(\displaystyle \frac{\partial f}{\partial x_{n}}=\frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}\)

this gives us the solution .

We note first that :

\(\displaystyle \frac{\partial f}{\partial x_{1}}=cos(x_{1}+2x_{2}+\cdots nx_{n})\)

Using the Chain Rule where \(\displaystyle x_{1}\) is the only variable here.

\(\displaystyle \frac{\partial f}{\partial x_{2}}=2 cos(x_{1}+2x_{2}+\cdots nx_{n})\)

Using the Chain Rule where \(\displaystyle x_{2}\) is the only variable here.

Continuing in this fashion we have:

\(\displaystyle \frac{\partial f}{\partial x_{n}}=ncos(x_{1}+2x_{2}+\cdots nx_{n})\)

Again using the Chain Rule and assuming that \(\displaystyle x_{n}\) is the variable and all the others are constant.

Now applying the given definition of the gradient we have the required result.

 \(\displaystyle [cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]\)

To find the derivative of this vector, all we need to do is to differentiate each component with respect to t.

Use the Power Rule and the Chain Rule when differentiating.

\(\displaystyle (\sqrt{1+t^2})dt=(1+t^2)^\frac{1}{2}dt=\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{t}{\sqrt{t^2+1}}\)  is the derivative of the first component.

\(\displaystyle (2\sqrt{1+t^2})dt=2(1+t^2)^\frac{1}{2}dt=2\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{2t}{\sqrt{t^2+1}}\) of the second component.

\(\displaystyle \vdots\)

\(\displaystyle (n\sqrt{1+t^2})dt=n(1+t^2)^\frac{1}{2}dt=n\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{nt}{\sqrt{t^2+1}}\) is the derivative of the last component . we obtain then:

\(\displaystyle \left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)\)

In general:

If \(\displaystyle \newline \ {\vec{r}(t)} = (f(t),g(t),z(t))\),

then \(\displaystyle \newline \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (\frac{\mathrm{d} }{\mathrm{d} t}f(t), \frac{\mathrm{d} }{\mathrm{d} t}g(t),\frac{\mathrm{d} }{\mathrm{d} t}z(t))\)

Derivative rules that will be needed here:

• Taking a derivative on a term, or using the power rule, can be done by doing the following:\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\)
• When taking derivatives of sums, evaluate with the sum rule which states that the derivative of the sum is the same as the sum of the derivative of each term: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(t^n+10) = \frac{\mathrm{d} }{\mathrm{d} t}t^n + \frac{\mathrm{d} }{\mathrm{d} t}10 = n*t^{n-1}\)
• Special rule when differentiating an exponential function: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} e^{kt} = ke^{kt}\) , where k is a constant.

In this problem, \(\displaystyle {\vec{r}(t)} = ((3e^t + t^3),(e^{2t} + t),(9e^t+e^{2t}))\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(3e^t + t^3)= \frac{\mathrm{d} }{\mathrm{d} t}3e^t + \frac{\mathrm{d} }{\mathrm{d} t}t^3 = 3e^t + 3t^2\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(e^{2t} + t)= \frac{\mathrm{d} }{\mathrm{d} t}e^{2t}+ \frac{\mathrm{d} }{\mathrm{d} t}t =2e^{2t} + 1\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}(9e^t+e^{2t})= \frac{\mathrm{d} }{\mathrm{d} t}9e^t + \frac{\mathrm{d} }{\mathrm{d} t}e^{2t} = 9e^t + 2e^{2t}\)

Put it all together to get \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t}{\vec{r}(t)} = (3e^t + 3t^2,2e^{2t} + 1,9e^t + 2e^{2t})\)