# Differential Calculus Problems

1. Mar 28, 2008

### RedBarchetta

Any help is greatly appreciated!!

1. The problem statement, all variables and given/known data

1.Differentiate.
http://www.webassign.net/www16/symImages/8/a/e5af282af9dd30006849e16c0b489b.gif

2.Find the derivative of the function.
http://www.webassign.net/www16/symImages/c/8/fd38a158e810bff80a28202fbceb37.gif

3.Consider the following.
http://www.webassign.net/www16/symImages/8/a/68e64f2907886f5478f5c03714da01.gif
(a) Find y ' by implicit differentiation.
(b) Solve the equation explicitly for y and differentiate to get y ' in terms of x.

4.Differentiate.
http://www.webassign.net/www16/symImages/1/a/d96706921943b5f83c856dd9fda12e.gif

5.Differentiate
http://www.webassign.net/www16/symImages/d/d/cc0e49fd81af528239a0b21a0ac5e9.gif

3. The attempt at a solution

1.On this one I started with the Quotient Rule.
y=sinx/1+cosx
y'=cosx(1+cosx)+(sinx)^2/(1+cosx)^2
y'=cosx+(cosx)^2+(sinx)^2/(1+cosx)^2
I type this in:
(cos(x)+cos^2(x)+sin^2(x))/(1+cos^2(x))
It says: "Check the syntax of your response." I've tried putting them like(cos(x))^2 and (cos^2(x)).

2.I tried this one with the chain rule.
I converted it:
s(t)=(t^3+3/t^3-3)^1/4
Now the derivative of the inside is what gets me.
s'(t)=1/4(t^3+3/t^3-3)^-3/4*(inside)
The inside derivative ends up equaling zero. I'm stumped. I used the quotient rule. Heres my attempt of the inside derivative=
3t^2(t^3-3)-(t^3+3)3t^2/(t^3-3)^2
The numerator equals zero, so that times the first part is zero, leaving me with zero. I tried zero but it was wrong.

3.Alright, differentiate with respect to x right? I started by converting the radicals.
x^1/2+y^1/2=6
Derivative:
1/2x^(-1/2)+1/2y^(-1/2)y'=6
6 is a constant and equals zero, so I moved the x term over, then divided by 1/2y to get y' by it self:
(-.5x^(.5))/(.5y^(-.5))
Wrong. I didnt attempt the second part yet.

4. The derivative(so i've read) of e^x, is simply e^x. I converted the radical. Then applied the product rule.
1/2x^(-1/2)*e^x+x^(1/2)*e^x
Wrong. I think I might have not applied the chain rule to the last term. so If you take the derivative of e^x, it should be e^x? But that shouldn't matter because its still e^x. Right?

5. I took the 2nd power and applied to the the entire term.
(1-u/1+u)^2
Then used chain rule.
2(1-u/1+u)*(inside derivative)
inside=
-1(1+u)-(1-u)*1/(1+u)^2
-1-u-1+u/(1+u)^2
-2/(1+u)^2
2(1-u/1+u)*(-2/(1+u)^2)
Still wrong. I'm not sure. This is what I typed: 2((1-u)/(1+u))*(-2)/(1+u)^2

Last edited by a moderator: Apr 23, 2017
2. Mar 28, 2008

### cristo

Staff Emeritus
This looks correct, but you should simplify it: use the fact that sin^2x+cos^2x=1. You should then get an expression that you can simplify further by factorising the denominator.

The numerator of the "inside" derivative does not equal zero: 3t^2(t^3-3)-(t^3+3)3t^2=-9t^2-9t^2=-18t^2
You've got $$\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{y}}y'=0$$ Try to make y' the subject of the formula again-- you're algebra went wrong somewhere.

Why's this wrong? Looks fine to me!
That's not correct since $$\frac{1-u^2}{1+u^2}\neq\left(\frac{1-u}{1+u}\right)^2$$.

You should just be able to use the quotient rule on the original expression.

3. Mar 28, 2008

### HallsofIvy

Staff Emeritus
I hate automated "homework checkers"! What you have for 4 is correct. But since the original problem was given in terms of $\sqrt{x}$, not x1/2, it would be better to write your answer in that form: What you have is the same as ((1/2)x-1/2+ x1/2)ex. If you factor out a (1/2)x-1/2, that is the same as (1/2)x-1/2(1+ 2x)ex which can be written
$$\frac{1+ 2x}{2\sqrt{x}}e^x$$

4. Mar 28, 2008

### RedBarchetta

On number 1 I got to this:
(cos(x))/(cos^2(x)+2cos(x))

Still says check your syntax.

I'm still stumped on number 2 and number 5.

Any help is appreciated.

Thanks!

5. Mar 29, 2008

### Snazzy

1.$$(1+cos(x))^2 \neq 1+cos^2x$$

2.$$\frac{(t^3+3)^\frac{1}{4}}{(t^3-3)^\frac{1}{4}}$$
Take it slow, apply the chain rule and the quotient rule.

3. Again, take it slow and apply the chain rule and the quotient rule. Make sure to be careful with your work.

Last edited: Mar 29, 2008