How do I differentiate the square root of x?
[Here, w.r.t means with respect to.]You mean differentiating root x w.r.t x? That's nothing but differentiating x^(1/2) w.r.t x. When you do that, you reduce the power of x by 1 and multiply the result by 1/2. Reducing power of x by 1, we get x^(-1/2), which is just the reciprocal of x^(1/2). So, the answer is 1/(2*x^1/2).
How to Differentiate 2x/square root(2x-1)?
.........2x y = ----------- .....√(2x - 1) y = 2x(2x - 1)^(-1/2) dy/dx = (2x - 1)^(-1/2) d/dx 2 + 2x d/dx - 1/2(2x - 1)^(-3/2) d/dx 2 ..........2.................2x =[ ------------------ - -----------------] √(2x - 1)^3 ....√(2x - 1).........√(2x - 1)^3 ......2√(2x - 1)^2 - 2x =---------------------------------- .........√(2x - 1)^3 .....2(2x - 1) - 2x =------------------------- .......√(2x - 1)^3 ......4x - 2 - 2x =---------------------- ...(2x - 1)^(3/2) ......2x - 2 =----------------- ...(2x - 1)^(3/2) ....2(x - 1) =----------------- answer// . (2x - 1)^(3/2)
Differentiate y = sin sqrt(2x)?
y = sin sqrt(2x) y' = cos sqrt(2x) d/dx( (sqrt(2x) ) y' = cos sqrt(2x) (1 /sqrt(2x)) y' = cos sqrt(2x) / sqrt(2x) Note: The derivative of sqrt(2x) = d/dx (sqrt(2x)) = d/dx (2x)^(1/2) = (1/2)(2x)^(-1/2) (2) = (2x)^(-1/2) = 1/(2x)^(1/2) = 1/sqrt(2x)
Differentiate sqrt(2x) - sqrt(3x^2)?
I just worked this problem on paper. It will be impossible to explain this under Yahoo Answers format. You will have to use the chain rule to solve this. That's the easy part. The hard part is simplying the expression after the chain rule is used. I wish I could explain this in person because it will be impossible to understand this over the computer. Before you can use the chain rule, you will have to rewrite the square root of each term, as each term raised to the (1/2) power √(2x) - √(3x²): before rewriting the square root as raised to the (1/2) power (2x)^(½) - (3x²)^(½): after rewriting the square root as raised to the (1/2) power From here, use the chain rule. (which is the easy part) The hard part is rationalizing the denominator and simplifying down to the final answer. This problem is too complex to understand over the internet. I recommend you see your instructor or find a personal tutor. I wish you the best of luck
Sin⁻¹(2x*sqrt(1-x^2)). what is the differentiation. if we let x=sin z then the answer will be one type?
sin⁻¹(2x*sqrt(1-x^2)). what is the differentiation. if we let x=sin z then the answer will be one type or if we assume x= cos z then the answer comes another type. are they both right? if we assume x= sin z sin⁻¹(2sin z*sqrt(1-sin^2 z)) =sin⁻¹(2sin z cos z ) =sin⁻¹(sin 2z) =2z =2 sin⁻¹x if we assume x= cos z sin⁻¹(2cos z*sqrt(1-cos^2 z)) =sin⁻¹(2cos z sin z ) =sin⁻¹(sin 2z) =2z =2 cos⁻¹x now if we differentiate in both cases the answer does not match. there comes a - sign in last case . so which one is correct? or both correct?