find the first derivative of the following functions:
y=[((x^2)-4)/(x+2)]
g(x)=[(x+1)/(x^1/2)]
Let's make like easier by doing some algebra first.
y=((x^2)-4)/(x+2) = x-2
dy/dx = 1
g(x)=[(x+1)/(x^1/2)]
= x^(1/2) - x^-1/2)
dg/dx = (1/2) x^-1/2 + (1/2)x^-3/2
To find the first derivative of a function, you can use the power rule, quotient rule, or chain rule, depending on the complexity of the function. In this case, we need to use the quotient rule for the first derivative of y=((x^2)-4)/(x+2) and the power rule for the first derivative of g(x)=[(x+1)/(x^1/2)].
For y=((x^2)-4)/(x+2), let's apply the quotient rule:
First, differentiate the numerator: d/dx(x^2 - 4) = 2x.
Then, differentiate the denominator: d/dx (x + 2) = 1.
Now apply the quotient rule:
(dy/dx)= (2x * (x+2) - (x^2 - 4) * 1) / (x+2)^2.
Simplifying the equation, we have:
(dy/dx) = (2x^2 + 4x - (x^2 - 4)) / (x+2)^2
(dy/dx) = (x^2 + 4x + 4) / (x+2)^2
(dy/dx) = (x+2)(x+2) / (x+2)^2
Since (x+2)^2 cancels out, the first derivative of y=((x^2)-4)/(x+2) is dy/dx = (x+2)/(x+2) = 1.
For g(x)=[(x+1)/(x^1/2)], let's apply the power rule:
First, simplify the expression g(x):
g(x) = x^(1/2) - x^(-1/2).
To find the first derivative, apply the power rule for each term:
For the first term, (d/dx)(x^(1/2)) = (1/2) * x^(-1/2) = 1/(2 * √(x)).
For the second term, (d/dx)(x^(-1/2)) = -(1/2) * x^(-3/2) = -1/(2 * x^(3/2)).
Combining the two terms, we have the first derivative of g(x):
(dg/dx) = (1/(2 * √(x))) - (1/(2 * x^(3/2))).
So, the first derivative of g(x)=[(x+1)/(x^1/2)] is dg/dx = (1/(2 * √(x))) - (1/(2 * x^(3/2))).