You can easily derive it from formula for the derivative of a power, if you remember that the cubic root of x is equal to x1/3.
This question asks for the proof of the derivative, not the derivative itself.
Using the definition of derivative, lim f(x) as h approaches 0 where f(x) = (f(a+h)-f(a))/h, we get the following:
[(a+h)1/3 - a1/3]/h
Complete the cube with (a2 + ab + b2)
Multiply by [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3] / [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3]
This completes the cube in the numerator, resulting in the following:
(a + h - a) / (h × [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3])
h / (h × [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3])
h cancels
1 / [(a+h)2/3 + (a+h)1/3 × a1/3 + a2/3]
Now that we have a function that is continuous for all h, we can evaluate the limit by plugging in 0 for h.
This gives
1/[a2/3 + a1/3 × a1/3 + a2/3]
Simplify a1/3 × a1/3
1/[a2/3 + a2/3 + a2/3]
(1/3)a2/3 or (1/3)a-2/3
This agrees with the Power Rule.
f'(x)=.5x^-.5 f''(x)=-.25x^-1.5 f'''(x)=.375x^-2.5 or f'''(x)=(3/8)x^(-5/2)
Derivative of lnx= (1/x)*(derivative of x) example: Find derivative of ln2x d(ln2x)/dx = (1/2x)*d(2x)/dx = (1/2x)*2===>1/x When the problem is like ln2x^2 or ln-square root of x...., the answer won't come out in form of 1/x.
0
It is x to the sixth power.
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
The derivative of cos x is -sin x, the derivative of square root of x is 1/(2 root(x)). Applying the chain rule, the derivative of cos root(x) is -sin x times 1/(2 root(x)), or - sin x / (2 root x).
Oh, dude, the third derivative of ln(x) is -2/(x^3). But like, who really needs to know that, right? I mean, unless you're planning on impressing your calculus teacher or something. Just remember, math is like a puzzle, except no one actually wants to put it together.
The square root of x = x to the power of a half
Use the formula for the derivative of a power. The square root of (x-5) is the same as (x-5)1/2.
If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)If you mean:f(x) = x1 + root(2)The derivative of x1, or x, is simply 1. The derivative of the square root of 2, just like the derivative of any constant, is zero. Therefore, the derivative of the entire function is one.If you mean:f(x) = x1 + root(2)you shuld use the power rule (the exponent, multiplied by x to the power (exponent minus 1)):(1 + root(2)) xroot(2)
1 divided by x to the third power equals x to the negative third. The derivative of x to the negative third is minus three x to the negative fourth.
1/2rootx
The derivative, with respect to x, is -x/sqrt(1-x2)
x*x1/2= x3/2 Derivative = 3/2 * x1/2
f'(x)=.5x^-.5 f''(x)=-.25x^-1.5 f'''(x)=.375x^-2.5 or f'''(x)=(3/8)x^(-5/2)
3/(4*square root(x)) ....Mukesh
hello how r u