Derivative of ln sqrt((x-1)/(x+1))?
y = ln sqrt((x-1)/(x+1)) y = (1/2)ln(x - 1) - (1/2)ln(x + 1) dy/dx = (1/2)(1/(x - 1)) - (1/2)(1/(x + 1)) dy/dx = [(x + 1) - (x - 1)] / [2(x - 1)(x + 1)] dy/dx = 2 / [2(x² - 1)] dy/dx = 1 / (x² - 1)
Derivative of ln sqrt(x^2+4)?
= x/sqrt(x^2+4) Use the method derivative of sqrt(f(x)) = 1/(2sqrt(f(x)) df(x)/dx Here f(x) =x^2+4.
What is the derivative of ln (3x^4 + 4x^3 + 3x^2 + x) (sqrt {x} + 2x)?
I understood this question as (ln(3x^4 + 4x^3 + 3x^2 + x))*( sqrt(x) + 2x).Begin by doing product rule, so derivative of first times second + derivative of second times first, so:(1/(3x^4 +4x^3 + 3x^2 + x)(12x^3 + 12x^2 + 6x + 1)(sqrt(x) + 2x) + (ln(3x^4 + 4x^3 +3x^2 + x))(1/(2sqrt(x)) +2)
Derivative of (x)/sqrt(x^2-4)?
Use the quotient rule. f(x) = x/√(x^2-4) d[√(x^2-4)]/dx = (1/2)(x^2 - 4)^(-1/2)(2x) = x/√(x^2 - 4) f'(x) = [x/√(x^2-4) - x^2/√(x^2 - 4)]/(x^2-4) f'(x) = [(x - x^2)/√(x^2-4)]/(x^2-4) Somehow this can be reduced to 1/√(x^2 - 4) - (x^2)/(x^2 - 4)^(3/2) I'm not sure how, but I don't think that much simplification is necessary. f'(x) = (x - x^2)/(x^2 - 4)^(3/2)
What is the derivative of x/ sqrt (x^2 +1)?
Applying the Quotient Rule to x/sqrt(x^2+1) gives: (sqrt(x^2+1) - x (1/(2 sqrt(x^2+1))) (2x))/(x^2+1) Now separate the two derivations on the top to get: sqrt(x^2+1)/(x^2+1) - (x (1/(2 sqrt(x^2+1))) (2x))/(x^2+1) Notice that the fraction on the left simplifies to: 1/sqrt(x^2+1) We are now done with the first fraction. Combining like terms on the second fraction yields: ((2 x^2)/(2 sqrt(x^2+1)))/(x^2+1) Now factor out the 2: ((x^2)/(sqrt(x^2+1)))/(x^2+1) What we now have is a fraction being divided by a quantity. So we'll multiply the two denominators together: x^2/(sqrt(x^2+1) (x^2+1)) Simplifying gives: x^2/(x^2+1)^(3/2) So the final answer is: 1/sqrt(x^2+1) - x^2/(x^2+1)^(3/2) Hope that helps.
Derivative of ln (x/sqrt(x^2+1))?
Basically you have to use the chain rule for the ln, the quotient rule for what's inside, and the chain rule again when taking the derivative of the denominator. You can simplify it more if you would like Correction: 1 / (x(x^2 + 1))
Derivative of ln sqrt x^2-4?
The derivative of the natural log is very simple: d/dx [ lnx ] = 1 / x You also need to know three other very simple rules: d/dx [ sqrtx ] = 1/(2sqrtx) d/dx [ x^n ] = nx^(n-1) d/dx [ c ] = 0 We need to complete this by doing the chain rule. y = ln(sqrt[x^2 - 4]) y' = (1 / sqrt(x^2 - 4)) * d/dx [sqrt(x^2 - 4)] y' = (1 / sqrt(x^2 - 4)) * (1/2sqrt(x^2 - 4)) * d/dx[x^2 - 4] y' = (1 / sqrt(x^2 - 4)) * (1/2sqrt(x^2 - 4)) * 2x y' = 2x / [2 * sqrt(x^2 - 4) * sqrt(x^2 - 4) ] y' = x / [ sqrt(x^2-4) * sqrt(x^2-4) ] y' = x / (x^2 - 4)
What is the derivative of [math]y=x\sqrt{1-x^2}[/math]?
If y = xsqrt(1 - x^2) then y^2 = x^2(1 - x^2) = x^2 - x^4. Now differentiate implicitly:2ydy = (2x - 4x^3)dx, therefore, cancel 2s and x and make dy/dx the subject of the formula:ydy = (x - 2x^3)dx = x(1 - 2x^2)dx=> dy/dx = x(1 - 2x^2)/y=> dy/dx = x(1 - 2x^2)/[xsqrt(1 - x^2)]= (1 - 2x^2)/[sqrt(1 - x^2)]
What's the derivative of y= (sqrt x) + 1/3sqrt x^4?
The simple differentiation will lead us to the answer but the domain and range is very important.domain of given function is x>=0.and range is also from [0,infinity) in case you are confusing to convert sqrtx^4=x^2.The derivative will be -1/2.x^(-1/2)+(2/3)x. Where x is non negative.