I have 3 calc problems that i have not been able to do please help me... i've been on here yesterday as well but no one was able to assist me with these particular ones...
1. If f(x)=5x^3–2/x^4
find f'(x)____
what i got was (4x^2+7x-12)/x^8
2. The position function of a particle is given by s=t^3–4t^2–7t t>=0
where s is measured in meters and t in seconds. When does the particle reach a velocity of 10 m/s? ____
3. f(x)=2x^2+7x+3/ sqroot of x
Find f'(x)
i got ((x^(-1/2))(4x+7)+1/2x^(-3/2)(2x^2+7x+3))/x
i had asked earlier and Damon answered them but it turned out to be wrong
I assume you are referring to
http://www.jiskha.com/display.cgi?id=1209342655http://www.jiskha.com/
Damon answered your #1 question correctly according to the way you typed it, but now your denominator is different, so which way is it?
If you want it the way you typed it this time, f(x) = If f(x)=5x^3–2/x^4
then f'(x) = [x^4(15x^2) - (5x^3-2)(4x^3)]/x^8
take out a common factor of x^3 from the top
= x^3(15x^3 - 4(5x^3-2))/x^8
= (-5x^2+8)/x^5
another way to do this is to rewrite the original
If f(x)=5x^3–2/x^4 as
f(x) = 5x^-1 - 2x^-4
then
f'(x) = -5x^-2 + 8x^-5
= -5/x^2 + 8/x^5
getting a common denominator and simplifying gives you the same answer of
(-5x^3+8)/x^5
In the second question Damon made a typo by setting the derivative equal to 7 instead of 10.
If you had looked at his solution you should have noticed that, and should have been able to make the necessary changes.
for your third I suggest you write it as
f(x)=2x^2+7x+3/ sqroot of x
= f(x)=2x^2+7x+3/ x^(1/2) then do what I did in #1
so f(x) = 2x^(3/2) + 7x^(1/2) + 3x^(-1/2)
f'(x) = 3x^(1/2) + (7/2)x^(-1/2) - (3/2)x^(-3/2)
I see (1/2)x^(-3/2) as a common factor
so
f'(x) = (1/2)x^(-3/2)[6x^2 + 7x - 3]
= (6x^2 + 7x - 3)/(2x^(3/2))
I'll be happy to help you with your calculus problems. Let's go through each problem step by step:
1. To find the derivative of f(x) = 5x^3 – 2/x^4, we can use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), then the derivative f'(x) can be found as [g'(x)h(x) - g(x)h'(x)] / [h(x)]^2.
In this case, g(x) = 5x^3 – 2 and h(x) = x^4. Let's differentiate each part:
g'(x) = 3 * 5x^2 - 0 (since there is no x term in -2)
= 15x^2
h'(x) = 4x^3 (we differentiate x^4 using the power rule)
Now, applying the quotient rule:
f'(x) = [g'(x)h(x) - g(x)h'(x)] / [h(x)]^2
= [(15x^2)(x^4) - (5x^3 – 2)(4x^3)] / (x^4)^2
= (15x^6 - (20x^6 – 8x^3)) / x^8
= (-5x^6 + 8x^3) / x^8
= -5/x^2 + 8/x^5
So, the derivative of f(x) is f'(x) = -5/x^2 + 8/x^5.
2. To find when the particle reaches a velocity of 10 m/s, we need to find the time value (t) when the velocity (v) is 10 m/s. Remember that velocity is the derivative of position with respect to time.
Given: s(t) = t^3 – 4t^2 – 7t
We need to find t when v(t) = s'(t) = 10. To find the derivative, we use the power rule:
s'(t) = 3t^2 – 8t – 7
Now we can set s'(t) = 10 and solve for t:
3t^2 – 8t – 7 = 10
Rearranging this equation, we get:
3t^2 – 8t – 17 = 0
Using the quadratic formula, t can be found:
t = (-(-8) ± √((-8)^2 - 4(3)(-17))) / (2(3))
Simplifying this expression gives us two possible values of t. You can evaluate these values using a calculator or follow the steps to solve the quadratic equation.
3. To find the derivative of f(x) = (2x^2 + 7x + 3) / sqrt(x), we need to use the quotient rule again.
Let g(x) = 2x^2 + 7x + 3 and h(x) = sqrt(x). Let's differentiate each part:
g'(x) = 4x + 7 (using the power rule for x^2)
h'(x) = (1/2) * x^(-1/2) (using the power rule for sqrt(x))
Now, applying the quotient rule:
f'(x) = [g'(x)h(x) - g(x)h'(x)] / [h(x)]^2
= [(4x + 7)sqrt(x) - (2x^2 + 7x + 3)(1/2)x^(-1/2)] / (sqrt(x))^2
= [(4x + 7)sqrt(x) - (2x^2 + 7x + 3)(1/2)x^(-1/2)] / x
= [2(2x + 7)sqrt(x) - (2x^2 + 7x + 3)/sqrt(x)] / x
= [2(2x + 7)sqrt(x) - (2x^2 + 7x + 3) / sqrt(x)] / x
So, the derivative of f(x) is f'(x) = [2(2x + 7)sqrt(x) - (2x^2 + 7x + 3) / sqrt(x)] / x.
I hope this explanation helps you understand how to solve these calculus problems!