Help with some limit and derivative problems

[djstealth]

Hello everyone,

I am having issues with some calculus problems. I have solved them all but I don't know if they are solved correctly so I would really appreciate it if you guys could check my work and maybe provide some insight on how to solve the problems if they are wrong.

The problems and my answers are:
1. Find the limit if it exists, if it doesn't explain why:
lim (2x+4) / |x + 2|
x->-2

my answer is undefined as the absolute value of x + 2 when x = -2 is 0

2. Find the derivative of y = cubic root of(x^4) + 1/(quad root of(x)); and simplify.

3*(x ^4/2) + x^-1/4
3*(2x) + (-1/4*x ^((-1/4)-1))
6x - 1/4*x^(-5/4)
6x - (1/4(x^5/4) < final answer


5. Find the derivative of (x)^2 / (2x^2 - x + 5) and simplify.

(2x^2 - x +5 * d/dx (x^2) - x^2 * d/dx (2x^2 - x +5)) / (2x^2 - x + 5)^2
(4x^3 - x^2 + 10x - 4x^3 + x^2) / (2x^2 - x + 5)^2
(-x^2 + 10x) / (2x^2 - x + 5)^2 < final answer


Thank you for your time.

 

[galactus]

1 and 3 are correct. But I do not know what 2 is supposed to be.

 

[djstealth]

I'm sorry for the confusion. I don't know how to post roots.

number 2 is as follows:

2. Find the derivative of y = cubic root of x^4 + 1/(quad root of x); and simplify.

3*(x ^4/2) + x^-1/4
3*(2x) + (-1/4*x ^((-1/4)-1))
6x - 1/4*x^(-5/4)
6x - (1/4(x^5/4) < final answer

 

[galactus]

As posted you have \(\displaystyle \sqrt[3]{x^{4}}+\frac{1}{\sqrt[4]{x}}=x^{\frac{2}{3}}+x^{\frac{-1}{4}}\)

The derivative would be \(\displaystyle \frac{16x^{\frac{19}{12}}-3}{12x^{\frac{5}{4}}}\)

I reckon quad root means 4th root?.

 

[djstealth]

yes, that is the equation. I am going to try to get to your answer. Maybe I did something wrong along the way