basic calculus problems

**philipsk** · January 18th 2007, 12:42 PM

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.

**ThePerfectHacker** · January 18th 2007, 12:51 PM

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

**CaptainBlack** · January 18th 2007, 01:26 PM

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

**ThePerfectHacker** · January 18th 2007, 01:28 PM

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.

**CaptainBlack** · January 18th 2007, 01:48 PM

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

)

**topsquark** · January 18th 2007, 01:56 PM

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!

**topsquark** · January 18th 2007, 02:09 PM

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

**galactus** · January 18th 2007, 02:31 PM

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.

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