Find the derivative of the function y=x2 + 2x + 1/ square root of x2 + 2x -1
To find the derivative of the function y = x^2 + 2x + 1 / sqrt(x^2 + 2x - 1), we can use the quotient rule. The quotient rule states that if you have a function in the form f(x) = g(x) / h(x), where g(x) and h(x) are both differentiable functions, then the derivative of f(x) is given by:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
For our function, let's define g(x) = x^2 + 2x + 1 and h(x) = sqrt(x^2 + 2x - 1). Now we need to find the derivatives of g(x) and h(x).
1. Derivative of g(x):
To find g'(x), we take the derivative of each term separately:
g'(x) = d/dx (x^2 + 2x + 1)
= 2x + 2 (using the power rule for derivatives)
2. Derivative of h(x):
To find h'(x), we need to use the chain rule because h(x) is a composition of functions. Let's define a new function u(x) = x^2 + 2x - 1, then we have:
h(x) = sqrt(u(x))
Now we can find h'(x) using the chain rule:
h'(x) = d/dx [sqrt(u(x))]
= (1 / (2 * sqrt(u(x)))) * d/dx [u(x)]
= (1 / (2 * sqrt(x^2 + 2x - 1))) * d/dx (x^2 + 2x - 1)
= (1 / (2 * sqrt(x^2 + 2x - 1))) * (2x + 2)
= (x + 1) / sqrt(x^2 + 2x - 1)
Now that we have g'(x) and h'(x), we can plug them into the quotient rule formula:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
Substituting the values we found:
f'(x) = [sqrt(x^2 + 2x - 1) * (2x + 2) - (x^2 + 2x + 1) * (x + 1)] / (sqrt(x^2 + 2x - 1))^2
Simplifying this expression would give you the derivative of the function y = x^2 + 2x + 1 / sqrt(x^2 + 2x - 1).