Find dy/dx in the following:
y=[x/√(1-x²)]- sin‾¹x
y'={[(1-x²)^(1/2)-(-2x)(x)(1/2)(1-x²)^(-1/2)]/(1-x²)} - (1/√(1-x²))
Then I multiplied (-2x)(x)(1/2) together. How do I simplify the rest?
I agree so far with
y' = {[(1-x²)^(1/2)-(-2x)(x)(1/2)(1-x²)^(-1/2)]/(1-x²)} - (1/√(1-x²))
= {[(1-x²)^(1/2)-(-x^2(1-x²)^(-1/2)]/(1-x²)} - (1-x^2)^(1/2)/(1-x²)
I formed a common denominator for the last term
= (1-x^2)^(-1/2) [1-x^2 - x^2 - (1-x^2)]
= -(1+x^2)/(1-x^2)^(3/2)
There would be other variations, the trick is to recognize if they are the same
Thank you for the help!
To simplify the expression, we can start by combining like terms in the numerator:
[ (1-x²)^(1/2) - (-2x)(x)(1/2)(1-x²)^(-1/2) ] / (1-x²)
Simplifying the numerator further, we can distribute the terms inside the bracket:
= [ (1-x²)^(1/2) + 2x^2(1/2)(1-x²)^(-1/2) ] / (1-x²)
Next, we can simplify the second term in the numerator:
= [ (1-x²)^(1/2) + x^2(1-x²)^(-1/2) ] / (1-x²)
Now, we can combine the two terms in the numerator over a common denominator:
= [ (1-x²)^(1/2)(1-x²) + x^2 ] / (1-x²)
Expanding the numerator:
= [ (1-x^4)^(1/2) + x^2 ] / (1-x²)
Finally, we can simplify the expression by canceling out the common factor of (1-x²) in the numerator and denominator:
= (1-x^4)^(1/2) + x^2