Calculus Problem: derivatives... Please help?
I hope someone can help me with these equation problem...
Please find :
FIRST DERIVATIVE of the following:
1) y = a ^4 (a - x)^ -3
2) y = 6 √4+x
3) x = (3 - 2y)¾
SECOND DERIVATIVE of the following:
4) x = 1
--------------
√1 - t^2
5) x = (1 + t)^2
------------------
t^2
Please include your solution.
Thank you very much!!
Comments
Two Rules you need to do this
The Chain rule
As an example, x^2 + 1 is a function, so
y = (x^2 + 1)^3 is a function of a function
The derivative of a function of a function is
the derivative of the outer function multiplied by
the derivative of the inner function.
Now see how easy it really is.
Derivative of outer function:
Think of y = (x^2 + 1)^3 as (thing)^3 and treat thing as if it were just x
You know the derivative of x^3 is 3x^2, so write down 3(x^2 + 1)^2
Derivative of inner function:
1 is a constant with derivative zero, so the derivative of x^2 + 1
is the same as the derivative of x^2 which is just 2x
Now put this all together. If y = (x^2 + 1)^3
dy/dx = 3(x^2 + 1)^2 * 2x = 6x(x^2 + 1)^2
Soon you will get used to these and be able to see the answer more quickly.
If I was not clear, try reading it through again or you may prefer this page of explanation.
http://www.math.hmc.edu/calculus/tutorials/chainru...
The Product rule
This is even easier. If y = u *v (where u and v are functions of x)
y' = uv' + vu'
e.g if y = x^3*sin(x)
y' = x^3*cos(x) + 3x^2*sin(x)
We only need this rule in your last question
======================================
1) y = a^4 (a - x)^(-3) we only need the chain rule for this.
The variable is x, so notice that the a^4 part is constant and remains unchanged.
Outer: Think of (a - x)^(-3) as (thing)^(-3)
You already know the derivative of x^(-3) is -3x^(-4),
so write down -3(a - x)^(-4) for this part
Inner: a is a constant with derivative zero, so the derivative of a - x
is the same as the derivative of -x which is just -1
Put this all together
dy/dx = a^4 *[-3(a - x)^(-4)] * -1 = 3a^4 (a - x)^(-4) = 3a^4/(a - x)^(4)
2) I hope that I am correct in assuming this reads as
y = 6 sqrt(4+x) = 6 (4+x)^(1/2)
With less explanation but the same idea
dy/dx = 6*(1/2)*(4+x)^(-1/2)*1 =3/sqrt(4+x)
3) x = (3 - 2y)^(3/4)
It is not made clear whether you want dx/dy or dy/dx so I have done both.
Used same idea to get dx/dy (will use reciproxal to get dy/dx later).
dx/dy = (3/4)(3 - 2y)^(-1/4) * -2 = -3/[2(3 - 2y)^(1/4)]
dy/dx = (-2/3)(3 - 2y)^(1/4)
Although correct, this is in terms of y, and we prefer x
We can use [(3 - 2y)^(1/4)]^3 =x to see that
dy/dx = (-2/3)(3 - 2y)^(1/4) = (-2/3)x^(1/3)
[N.B. We could have got the same answer starting from
y = 3/2 - (1/2)x^(4/3) to get dy/dx ]
4) I am assuming this reads find second derivative of
x = (1 - t^2)^(-1/2)
x' = (-1/2) * (1 - t^2)^(-3/2) * -2t = t/(1 - t^2)^(3/2)
For x'', the next derivative, we need to use the product rule which
says that the derivative of u *v can be found from uv' + vu'
Let u = (1 - t^2)^(-3/2) and v = t, so that
x' = uv = t/(1 - t^2)^(3/2)
Then x'' will work out like u *1 + t *u'
We will have to work out u' using the earlier chain rule idea
u' = d/dt[(1 - t^2)^(-3/2) ] = (-3/2) (1 - t^2)^(-5/2) * -2t = 3t(1 - t^2)^(-5/2)
so x'' = u *1 + t *u' = (1 - t^2)^(-3/2) + t*[3t(1 - t^2)^(-5/2)]
x'' = [1 - t^2 + t*3t]]/(1 - t^2)^(5/2)
x'' = (1+2t^2 )/(1 - t^2)^(5/2)
5) I am assuming this reads find second derivative of
x = (1 + t^2)/t^2
If that is correct, it simplifies to t^(-2) +1
You might want to see if you can do this on your own.
If having trouble, post your efforts and/or questions.
I hope this was a helpful post for you.
Regards - Ian
1. a^4 is a constant coefficient which does not influence the derivative. We can just differentiate (a-x)^(-3) and multiply the result by a^4:
For derivative of (a-x)^(-3), use the rule (d/dx)(x^n) = nx^(n-1), and then the chain rule tells us to multiply by the derivative of a-x, that’s -1. Thus
(d/dx)((a-x)^(-3)) = -3(a-x)^(-4) * (-1)
............................ = 3(a-x)^(-4).
Multiplying by a^4, give the answer as 3(a^4)(a-x)^(-4) which can also be written as 3(a/(a-x))^4.
2) I guess the sq rt is of 4+x, not just 4, so we should write
y = 6â(4+x) = 6(4+x)^(1/2).
Multiply by (1/2) and then subtract 1 from the index ½:
dy/dx = 3(4+x)^(-1/2)
......... = 3/ â(4+x).
3) Multiply by ¾, subtract 1 from the index to get -¼, and the chain rule says to multiply by the derivative of 3-2y, i.e. -2, so we get
¾(3 - 2y)^(-¼) * -2
= -(3/2)(3-2y)^(-¼))
4) Again, I'm sure there should be parentheses around 1-t² to show that the sq rt doesn't apply only to 1:
x = 1/â(1-t²) = (1 - t²)^(-½)
Do you know the chain rule? Differentiate (1 - t²)^(-½) as something^(-½), getting -½ (something)^(-3/2), and then multiply by the derivative of 'something', which in this case is the derivative of 1-t², i.e. -2t:
dx/dt = -½ (1 - t²)^(-3/2) * (-2t)
........ = t (1 - t²)^(-3/2)
Now differentiate again using the product formula, u'v + uv'.
First differentiate t, giving 1; then for the next term differentiate (1 - t²)^(-3/2), giving
(-3/2)(1 - t²)^(-5/2) * (-2t), which of course still has to be multiplied by the other factor, t:
d²x/dt² = 1*(1 - t²)^(-3/2) + t* (3t)((1 - t²)^(-5/2)
........... = (1 - t²)^(-3/2) + 3t²(1 - t²)^(-5/2)
........... = (1 - t²)^(-5/2)[(1 - t²) + 3t²)
........... = (1 - t²)^(-5/2)(1 + 2t²)
5) I would simplify this first.
x = [(1+t)/t]²
.. = (1/t + 1)²
dx/dt = 2(1/t + 1) * (-1/t²) since derivative of t^(-1) is -t^(-2)
........ = -(2/t²)(1/t + 1)
........ = -2/t³ - 2/t²
........ = -2t^(-3) - 2t^(-4)
Hence
d²x/dt² = 6 t^(-4) + 8t^(-5)
.......... = 6/t⁴ + 8/t⁵
.......... = (6t + 8)/t⁵
EDIT COMMENT: It's always comforting to see another answerer gets the same results, as Ian does, with rather more thorough and systematic explanation. You may notice though that our results for q.5 are different, and that is because Ian has guessed that you mis-placed the parentheses, and actually meant (1 + t²)/t²