Differentiate the function.
g(u) = sqrt(2)u+sqrt(7u)
g'(u)=?
Can somebody get me started maybe with the first two steps and then from there Ill try finishing it and then come back on and post the answer I get.
when a is a constant, g'(au) = a g'
You know that if
y=2x, y'=2(1)
y=3x^7, y' = 3(7x^6)
The constant is just a multiplier, and carries through.
So, what you have is
g = √2 u + √7√u
To differentiate the function g(u) = sqrt(2)u + sqrt(7u), we need to find the derivative of each term separately and then add them together.
First, let's find the derivative of the first term sqrt(2)u. To do this, we can use the power rule for differentiation, which states that if we have a function of the form f(x) = x^n, then its derivative is f'(x) = nx^(n-1).
In this case, we have sqrt(2)u, which can be written as (2^(1/2))u. By applying the power rule, the derivative of this term is (1/2)(2^(1/2))u^(1-1) = (1/2)(2^(1/2))u^0 = (1/2)(2^(1/2)) = (2^(1/2))/2 = sqrt(2)/2.
Now, let's find the derivative of the second term sqrt(7u). Using the power rule, we have (7u)^(1/2). Again applying the power rule, the derivative of this term is (1/2)(7u)^(1/2 - 1) = (1/2)(7u)^(-1/2) = (1/2)(1/sqrt(7u)) = 1/(2sqrt(7u)).
Finally, we add the derivatives of both terms to get the derivative of the entire function g(u):
g'(u) = sqrt(2)/2 + 1/(2sqrt(7u))
Therefore, the derivative of the function g(u) is g'(u) = sqrt(2)/2 + 1/(2sqrt(7u)).