use implicit differentiation to show: d/dx(tan^-1x)=1/(x^2+1)
y = arctan(x)
tan(y) = x
sec^2(y) y' = 1
y' = 1/sec^2(y) = 1/(1+tan^2(y)) = 1/(1+x^2)
To use implicit differentiation to show that the derivative of arctan(x) with respect to x is 1/(x^2 + 1), we need to follow these steps:
Step 1: Start with the equation:
y = arctan(x)
Step 2: Apply the tangent function to both sides of the equation:
tan(y) = tan(arctan(x))
Step 3: Use the trigonometric identity tan(arctan(x)) = x:
tan(y) = x
Step 4: Take the derivative of both sides of the equation with respect to x:
d/dx(tan(y)) = d/dx(x)
Step 5: Apply the chain rule on the left side:
sec^2(y) * dy/dx = 1
Step 6: Solve for dy/dx, which is our desired result:
dy/dx = 1 / sec^2(y)
Step 7: Substitute the value of sec^2(y) using the Pythagorean identity:
dy/dx = 1 / (1 + tan^2(y))
Step 8: Replace tan(y) with x (from Step 3):
dy/dx = 1 / (1 + x^2)
Therefore, the derivative of arctan(x) with respect to x is 1/(x^2 + 1).