1. ## basic calculus problems

my son just asked me to help him with some calculus but i haven't done calculus in 30 years so i can't really help him... i was hoping that someone here could help. his homework is do tomorrow. so, time is essential.. well, i was hoping that someone could hopefully show me how to do these problems so i could show my son how to do them.

Thanks in advance to all that this may concern.

2. Originally Posted by philipsk
my son just asked me to help him with some calculus but i haven't done calculus in 30 years so i can't really help him...
Does he know that you are getting help or does he think that you know?

1) $f(x)=x-\sin 2x$
$f'(x)=(x)'-(\sin 2x)'=1-2\cos 2x$ (chain rule).

5) $y=\frac{\sin x}{4+\cos x}$
Quotient rule,
$y'=\frac{(\sin x)'(4+\cos x)-\sin x(4+\cos x)'}{(4+\cos x)^2}$
$y'=\frac{\cos x(4+\cos x)-\sin x(-\sin x)}{(4+\cos x)^2}$
$y'=\frac{4\cos x+\cos^2 x+\sin^2 x}{(4+\cos x)^2}=\frac{4\cos x+1}{(4+\cos x)^2}$

3. Originally Posted by ThePerfectHacker
Does he know that you are getting help or does he think that you know?
philipsk is a dad he is entitled to our help to appear omnicient in the eyes
of his chidren.

RonL

4. Originally Posted by CaptainBlack
philipsk is a dad he is entitled to our help to appear omnicient in the eyes
of his chidren.
I should have added that he should not say that he is getting help to his son, I forgot to.

5. 2. Find the equation of the tangent to the curve $y=\frac{\sqrt{x}}{x+6}$ at $(4,0.2)$

The equation of a line (and the tangent is a line) is of the form $y=mx+c$, and this is a tangent at the given point on the curve if $m$ is equal to the slope or gradient of the curve at the point and c is chosen so that the line passes through the point.

Now the slope of the curve at a point is equal to the derivative of $y$ with respect to [tex][x/math] at the point. To evaluate this derivative we need the quotient rule:

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2}$,

with $f(x)=\sqrt{x}$ and $g(x)=x+6$, so $f'(x)=\frac{1}{2 \sqrt{x}}$, and $g'(x)=1$

Hence:

$\frac{dy}{dx}=\frac{\frac{x+6}{2\sqrt{x}}-\sqrt{x}}{(x+6)^2}$

Now we could simplify this but I think it better to just evaluate it as it is at $x=4$, which I make: $\left. \frac{dy}{dx}\right|_{x=4}=0.005$.

So our line has the form: $y=0.005\,x+c$. Now when $x=4\ y=0.2$, so: $0.2=0.005\,4+c$, hence $c=0.18$, so finally we have the equation of the tangent at the given point is:

$y=0.005\,x+0.18$

RonL

(now I expect one of the other helpers will point out some dreadful mistake in my algebra )

6. 2. Find the equation of the line tangent to $y = \frac{\sqrt{x}}{x+6}$ at the point (4, 0.2).

The equation of any line may be expressed as
$y = mx + b$
where m is the slope of the line and b is the y-intercept (0, b).

The slope of the line tangent to $y = \frac{\sqrt{x}}{x+6}$ at the point (x, y) is the first derivative of y with respect to x. So by the quotient rule:
$y' = \frac{\frac{1}{2} \cdot \frac{1}{\sqrt{x}} \cdot (x + 6) - \sqrt{x} \cdot (1)}{(x+6)^2}$ <-- Mulitply top and bottom by $2\sqrt{x}$

$y' = \frac{x + 6 - 2x}{2 \sqrt{x} (x + 6)^2}$

$y' = \frac{-x + 6}{2 \sqrt{x} (x + 6)^2}$

So the slope of the tangent line to the curve at the point (4, 0.2) is:
$m = \frac{-4 + 6}{2 \sqrt{4} (4 + 6)^2} = 0.005$ <-- Or 1/200.

So the equation of the tangent line at (4, 0.2) is:
$y = 0.005x + b$

Now we need to find b, so put in the point coordinates:
$0.2 = 0.005 \cdot 4 + b$

$b = 0.18$ <-- Or 9/50

Thus the equation of the line is:
$y = \frac{1}{200}x + \frac{9}{50}$

I have attached a graph of the function and the tangent line.

-Dan

Originally Posted by CaptainBlack
(now I expect one of the other helpers will point out some dreadful mistake in my algebra )
At least we agree!

7. 3. Find the equation of the line tangent to $y = x^8cos(x)$ at $( \pi, -\pi^8)$.

Same method as before.

$y' = 8x^7cos(x) + x^8 \cdot - sin(x)$

$y' = -x^8sin(x) + 8x^7cos(x)$

At the point $( \pi, -\pi^8)$ this is:
$m = -(\pi)^8sin(\pi) + 8(\pi)^7cos(\pi) = 0 - 8\pi^7 = -8\pi^7$

So we have the tangent line:
$y = (-8\pi^7)x + b$
at the point $( \pi, -\pi^8)$. Again, we need b, so:

$-\pi^8 = (-8\pi^7) \cdot \pi + b$

$b = -\pi^8 + 8\pi^8 = 7\pi^8$

Thus the tangent line is:
$y = (-8\pi^7)x + 7\pi^8$

Again, I have attached a graph. (The scale of the y-axis hides some of the features of the graph.)

-Dan

8. For #7, polynomials are among the easiest functions to differentiate. They're everywhere continuous and very well-behaved.

$\frac{d}{dx}[2x^{3}-3x^{2}-36x+4]=6x^{2}-6x-36=x^{2}-x-6$

Factor:

$(x-3)(x+2)$

As you can see, you have zeros at x=3 and x=-2

See the graph. Note the horizontal tangents as the maximum and minimum.