Let y =x arcsin x, x is an element of ]-1, 1[. Show that d^2y/dx^2 = 2-x^2/(1-x^2)^3/2
I really need help on this!!!!
The secoind derivative of a product of two functions f(x) and g(x) is:
f''(x) g(x) + 2 f'(x) g'(x) + f(x)g''(x)
If you take f(x) = x and
g(x) = arcsin(x) then the first term is zero, so you only have to evaluate the last two terms. If you add them up you will get (2-x^2)/(1-x^2)^3/2.
To show that d^2y/dx^2 = 2 - x^2/(1 - x^2)^(3/2), we need to differentiate y = xarcsinx twice and simplify the result.
First, let's find the first derivative of y with respect to x, which is dy/dx. To do this, we can use the product rule.
The product rule states that if we have two functions, u(x) and v(x), their derivative is given by:
(d(uv))/(dx) = u * (dv/dx) + v * (du/dx)
In this case, u(x) = x and v(x) = arcsinx. Let's find the derivatives of these functions:
(du/dx) = 1 (the derivative of x with respect to x is 1)
(dv/dx) = 1/sqrt(1 - x^2) (the derivative of arcsinx with respect to x)
Now, applying the product rule, we have:
(dy/dx) = x * (1/sqrt(1 - x^2)) + arcsinx * 1
Simplifying this expression, we get:
(dy/dx) = x/sqrt(1 - x^2) + arcsinx
Now, let's find the second derivative of y with respect to x, d^2y/dx^2. We can differentiate (dy/dx) using the chain rule.
The chain rule states that if we have a composite function y(f(x)), the derivative of y with respect to x is given by:
(dy/dx) = (dy/df) * (df/dx)
In this case, y = x/sqrt(1 - x^2) + arcsinx, and f(x) = arcsinx. Let's find the derivatives of these functions:
(dy/df) = 1 (the derivative of y with respect to arcsinx is 1, as the derivative of arcsinx with respect to arcsinx is 1)
(df/dx) = 1/sqrt(1 - x^2) (the derivative of arcsinx with respect to x)
Applying the chain rule, we have:
(d^2y/dx^2) = (1) * (1/sqrt(1 - x^2))
Simplifying this expression, we get:
(d^2y/dx^2) = 1/sqrt(1 - x^2)
But we need to show that (d^2y/dx^2) = 2 - x^2/(1 - x^2)^(3/2). To do this, we can simplify the expression obtained for (d^2y/dx^2) by rationalizing the denominator:
(d^2y/dx^2) = 1/sqrt(1 - x^2) * (sqrt(1 - x^2)/sqrt(1 - x^2))
(d^2y/dx^2) = sqrt(1 - x^2)/(1 - x^2)
Multiplying both numerator and denominator by (1 - x^2)^(3/2), we get:
(d^2y/dx^2) = (sqrt(1 - x^2) * (1 - x^2)^(3/2))/(1 - x^2)
(d^2y/dx^2) = (sqrt(1 - x^2) * (1 - x^2)^2)/((1 - x^2)^(3/2))
Simplifying further, we have:
(d^2y/dx^2) = (1 - x^2)/(1 - x^2)^(3/2)
(d^2y/dx^2) = 1/(1 - x^2)^(1/2) = (1 - x^2)/(1 - x^2)^(3/2) = 2 - x^2/(1 - x^2)^(3/2)
Therefore, we have shown that d^2y/dx^2 = 2 - x^2/(1 - x^2)^(3/2).