y = arcsin(x/(sqrt(1 + x ^ 2))) + arccos(1/(sqrt(1 + x ^ 2))) find dy/dx
To find the derivative dy/dx, we first need to use the chain rule and differentiate each term separately.
Let's start with the first term y = arcsin(x/(sqrt(1 + x^2))).
Using the chain rule, we have:
dy/dx = (d/dx)arcsin(x/(sqrt(1 + x^2)))
Let u = x/(sqrt(1 + x^2)).
Differentiating u with respect to x, we have:
du/dx = (d/dx)x/(sqrt(1 + x^2))
To compute du/dx, we need to use the quotient rule:
(d/dx)x/(sqrt(1 + x^2)) = [(1)(sqrt(1 + x^2)) - x(1/2)(2x)] / (1 + x^2)
= (sqrt(1 + x^2) - x^2) / (1 + x^2)
= [(1 + x^2) - x^2] / (1 + x^2)
= 1 / (1 + x^2)
Now we can substitute du/dx into our original equation:
dy/dx = (d/dx)arcsin(u) = du/dx / sqrt(1 - u^2)
= 1 / (1 + x^2) / sqrt(1 - (x/(sqrt(1 + x^2)))^2)
= 1 / (1 + x^2) / sqrt(1 - (x^2 / (1 + x^2)))
= 1 / (1 + x^2) / sqrt((1 + x^2 - x^2) / (1 + x^2))
= 1 / (1 + x^2) / sqrt(1 / (1 + x^2))
= 1 / (1 + x^2) / (sqrt(1) / sqrt(1 + x^2))
= sqrt(1 + x^2) / (1 + x^2)
Now let's move on to the second term y = arccos(1/(sqrt(1 + x^2))).
Using the chain rule, we have:
dy/dx = (d/dx)arccos(1/(sqrt(1 + x^2)))
Let v = 1/(sqrt(1 + x^2)).
Differentiating v with respect to x, we have:
dv/dx = (d/dx)1/(sqrt(1 + x^2))
To compute dv/dx, we need to use the quotient rule:
(d/dx)1/(sqrt(1 + x^2)) = [(0)(sqrt(1 + x^2)) - 1(1/2)(2x)] / (1 + x^2)
= - x / (1 + x^2)^1.5
Now we can substitute dv/dx into our original equation:
dy/dx = (d/dx)arccos(v) = -dv/dx / sqrt(1 - v^2)
= -(- x / (1 + x^2)^1.5) / sqrt(1 - (1/(sqrt(1 + x^2)))^2)
= x / (1 + x^2)^1.5 / sqrt(1 - (1/(sqrt(1 + x^2)))^2)
= x / (1 + x^2)^1.5 / sqrt(1 - (1^2 / (1 + x^2)))
= x / (1 + x^2)^1.5 / sqrt((1 + x^2 - 1) / (1 + x^2))
= x / (1 + x^2)^1.5 / sqrt(x^2 / (1 + x^2))
= x / (1 + x^2)^1.5 / (sqrt(x^2) / sqrt(1 + x^2))
= x / (1 + x^2)^1.5 / (|x| / sqrt(1 + x^2))
= x / (1 + x^2)^1.5 / (|x| / (1 + x^2))
= x / (1 + x^2)^1.5 * ((1 + x^2) / |x|)
= x / |x|(1 + x^2)^(1.5)
Now we can simplify the expression for dy/dx as:
dy/dx = sqrt(1 + x^2) / (1 + x^2) - x / |x|(1 + x^2)^(1.5)
Simplify the expression further, we have:
dy/dx = (sqrt(1 + x^2) - x / |x|) / (1 + x^2)^(1.5)
Therefore, the derivative of y with respect to x, dy/dx, is given by:
dy/dx = (sqrt(1 + x^2) - x / |x|) / (1 + x^2)^(1.5)
To find dy/dx, we need to differentiate the given expression with respect to x.
Given: y = arcsin(x/(sqrt(1 + x^2))) + arccos(1/(sqrt(1 + x^2)))
First, let's differentiate the first term, which is arcsin(x/(sqrt(1 + x^2))).
We can use the chain rule to differentiate this term.
Let u = x / sqrt(1 + x^2), and y = arcsin(u).
Using the chain rule:
dy/du = 1 / sqrt(1 - u^2) (derivative of arcsin(u) with respect to u)
du/dx = (sqrt(1 + x^2) - x * (1 / (2 * sqrt(1 + x^2))) * (2 * x)) / (1 + x^2)
du/dx = (sqrt(1 + x^2) - x^2) / (sqrt(1 + x^2) * (1 + x^2))
Multiplying these two derivatives together using the chain rule:
dy/dx = (dy/du) * (du/dx)
dy/dx = (1 / sqrt(1 - u^2)) * ((sqrt(1 + x^2) - x^2) / (sqrt(1 + x^2) * (1 + x^2)))
Next, let's differentiate the second term, which is arccos(1/(sqrt(1 + x^2))).
We can use the chain rule here as well.
Let v = 1 / sqrt(1 + x^2), and z = arccos(v).
Using the chain rule:
dz/dv = -1 / sqrt(1 - v^2) (derivative of arccos(v) with respect to v)
dv/dx = (-2 * x) / (2 * sqrt(1 + x^2))^2
dv/dx = -2 * x / (4 * (1 + x^2))
Multiplying these two derivatives together using the chain rule:
dz/dx = (dz/dv) * (dv/dx)
dz/dx = (-1 / sqrt(1 - v^2)) * (-2 * x / (4 * (1 + x^2)))
dz/dx = (x / sqrt(1 - v^2)) / (2 * (1 + x^2))
Now, we can find dy/dx by summing the derivatives of the two terms:
dy/dx = (1 / sqrt(1 - u^2)) * ((sqrt(1 + x^2) - x^2) / (sqrt(1 + x^2) * (1 + x^2)))
+ (x / sqrt(1 - v^2)) / (2 * (1 + x^2))
Simplifying this expression further may be difficult, so this is the final result for dy/dx.