Find the derivative of the function.
f(x) = 7x^3 + 7x^2/x
f '(x) =
Please show us how you think these problems should be solved -- or where you're stuck. Do not post any more problems until you've shown us some effort on these other 10 or so problems.
7(3x^3-1)= 21x^2
7(2x^2-1)= 14x
so would it be 21x^2+14x/x
I don't know how to go further than this
I don't know the foggiest idea what you are doing.
first of all, I think you probably meant
f(x) = (7x^3 + 7x^2)/x , or else the last term would simply be 7x
from my assumption, then
f(x) = 7x^2 + 7x
f'(x) = 14x + 7
I think i see what i am doing wrong thank you very much :)
To find the derivative of the given function f(x) = 7x^3 + 7x^2/x, we can use the power rule and quotient rule.
Step 1: Apply the power rule to the first term.
The power rule states that the derivative of x^n is n*x^(n-1).
So, for the first term 7x^3, the derivative is:
d/dx (7x^3) = 3*7x^(3-1) = 21x^2.
Step 2: Apply the power rule to the second term.
For the second term 7x^2/x, we can rewrite it as 7x^(2-1)/x = 7x^1 = 7x.
Now, apply the power rule to get the derivative:
d/dx (7x) = 1*7 = 7.
Step 3: Apply the quotient rule to combine the derivatives of both terms.
The quotient rule states that for a function of the form f(x)/g(x), its derivative is (f'(x) * g(x) - f(x) * g'(x)) / (g(x))^2.
So, applying the quotient rule to the given function, we have:
f'(x) = (21x^2 * x - 7x^3 * 1) / (x)^2
= (21x^3 - 7x^3) / x^2
= 14x^3 / x^2
= 14x.
Therefore, the derivative of the function f(x) = 7x^3 + 7x^2/x is f'(x) = 14x.