Homework about Derivative.I know that It's quite easy but I need reassurance?

Question

Use these this rules in finding the derivative. Derivative of a constant, Derivative of a power, Derivative of a constant times a function, Derivative of sums and differences, Derivative of a product, and Derivative of a quotient.

Find the indicated derivative

1. y = 4x^5 – 3x^4 + 2x^3 – x + π, dy/dx
2. y = (3w^2 – 4)(3w^2 + 4), dy/dw
3. y = t^2 – 6t + 9/t^2 – 5t + 6, dy/dt
4. w = (t + 1)(t^2 – 1)/t^2 + 2t + 1, dw/dt
5. y = x^3 – 125/ x^2 + 5x + 25, dy/dx
6. y = (x / x+ 1 + x – 1 /x ), dy/dx
7. x = (t + 1)(t + 2)(t + 3), dx/dt
8. y = (1/x + 1/x^2 + 1/x^3), dy/dx

Please show a little of your solutions.

Answer 1

1)
y = 4x^5 – 3x^4 + 2x^3 – x + π
dy/dx = 4(5)x^4 - 3(4)x^3 + 2(3)x^2 - 1 + 0
= 20x^4 - 12x^3 + 6x^2 - 1

2)
y = (3w^2 - 4)(3w^2 + 4)
Use product rule:
U = 3w^2 - 4
U' = 3(2)w - 0 = 6w
V = 3w^2 + 4
V' = 3(2)w + 0 = 6w

dy/dw = U'V + UV'
= (6w)(3w^2 + 4) + (3w^2 - 4)(6w)
(you can mulitply it through if you want)

NOTE: you could also multiply through BEFORE taking the derivative, then it would just be a polynomial like #1

3)
y = (t^2 – 6t + 9)/(t^2 – 5t + 6)
Use quotient rule:
f(x) / g(x)
derivative = (f'g - g'f)/g^2

f(t) = t^2 - 6t + 9
f' = 2t - 6

g(t) = t^2 - 5t + 6
g' = 2t - 5

dy/dt = [(2t - 6)(t^2 - 5t + 6) - (2t - 5)(t^2 - 6t + 9)]
divided by (t^2 - 5t + 6)^2

4)
w = (t + 1)(t^2 - 1) / (t^2 + 2t + 1)
w = (t + 1)(t - 1)(t + 1) / (t + 1)(t + 1)
w = t - 1
so...
dw/dt = 1

5)
y = x^3 – 125/ x^2 + 5x + 25
y = (x - 5)(x^2 + 5x + 25) / (x^2 + 5x + 25)
y = x - 5

Now I think you can take the derivative yourself.

6)
y = [x /(x+1)] + [(x-1) / x]
It's probably easier to get a common denominator first.
y = [x(x) / x(x+1)] + [(x-1)(x+1) / x(x+1)]
y = [x^2 / (x^2 + x)] + [(x^2 - 1) / (x^2 + x)]
y = (2x^2 - 1) / (x^2 + x)

Now use quotient rule like I did in #3.

7)
There is a product rule for 3 terms, but I think it's probably easier to multiply two of the terms together, then use product rule.
x = (t + 1)(t + 2)(t + 3)
x = (t^2 + 3t + 2)(t + 3)

Now use product rule like in #2.

8)
y = (1/x) + (1/x^2) + (1/x^3)
y = x^(-1) + x^(-2) + x^(-3)

dy/dx = (-1)x^(-2) + (-2)x^(-3) + (-3)x^(-4)
= - (1 / x^2) - (2 / x^3) - (3 / x^4)

Answer 2

I use D as the Derivative function (easier to write than dy/dx).

Derivative of a constant, e.g., D(k) where k is a constant
always 0 (the "change" in value is 0)

Derivative of a power
D(x^p) where x is the variable and p the power
= p * x^(p-1)
the power becomes a factor, the new power is one less.

Derivative of a constant times a power
D(k*(x^p)) = k*D(x^p)
If a function varies according to x, it has a certain slope (same value as the derivative); if the function varies according to k*x (k being a factor), then its slope is k times steeper.

Derivative of sums = sums of Derivatives
D(f+g) = D(f) + D(g) where f and g are functions of x.
Derivative of difference = difference of Derivatives
D(f-g) = D(f)-D(g)

Derivative of product is slightly tricker:
D(fg) = f*D(g) + g*D(f)

Derivative of a quotient of functions, like f/g
1st, you must make sure that g cannot have a value of 0.
D(f/g) = [f*D(g) - g*D(f)] / g^2

1. uses the first three rules:
D(4x^5 – 3x^4 + 2x^3 – x + π) =
D(4x^5) - D(3x^4) + D(2x^3) - D(x) + D(pi)
x is the same as x^1; pi is a constant
4*D(x^5) - 3*D(x^4) + 2*D(x^3) - D(x^1) + D(constant)
4*(5x^4) - 3*(4x^3) + 2*(3x^2) - 1x^0 + 0
x^0 = 1
20x^4 -12x^3 +6x^2 -1

7. Product rule twice! (You could simply multiply it out in which case it becomes the same kind of problem as number 1, but that would be no fun: too easy)
D[(t+1)(t+2)(t+3)]
we will use f=(t+1) and g=(t+2)(t+3)
(t+1)*D[(t+2)(t+3)] + (t+2)(t+3)*D(t+1)
we use the product rule again to do D[(t+2)(t+3)]
(t+1)*{ (t+2)*D(t+3) + (t+3)*D(t+2) } + (t+2)(t+3)*D(t+1)
each of the D(t+a) is a sum of D(t) + D(constant) = 1 + 0.
(t+1)*{ (t+2)*1 + (t+3)*1 } + (t+2)(t+3)*1
(t+1)(t+2) + (t+1)(t+3) + (t+2)(t+3)
multiply it out to get
3t^2 + 12t + 11
although I prefer the one with the patterned products, because it shows:
Let f=(t+1), g=(t+2), h=(t+3)
Then D(f)=1, D(g)=1 and D(h)=1
D(fgh) = fg*D(h) + fh*D(g) + gh*D(f)
(which is the "Derivative of a product" extended to 3 factors).

Answer 3

1. 20 x^4 - 12 x^3 + 6x^2 -1
2. 36 w^3
3. 2t - 6 - 18/x^3 + 2
4. 1
5. 3x2 + 125/x^3 + 5
6. -1/(x+1)^2 + 1/x^2
7. 3x^2 + 12x +11
8. -1/x^2 - 2/x^3 -3/x^4

Answer 4

1. 20x^4 - 12x^3 + 6x^2 -1
2. 36w^3
3. 2t - 6 - 18/t^3 -5
4. 1
5. 3x^2 +250/x^3 +5
6. 1 + 1/x^2
7. (t+2)(t+3) + (t+1)(t+3) + (t+1)(t+2)
8. -1(1/x^2 + 2/x^3 + 3/x^4)
No.3 & 2 are quite easy once you simplify them

Answer 5

It is quite easy, but why not posts your solutions for someone to verify instead of asking people to do all the work for you?