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What is the linearization L of x of the function at a. If f of x equals x to the 4 power plus 3x square a equals negative 1?
the linearization of f(x) = x^4 + 3x^2 with a = -1 is L(x) = 4 - 10(x+1) the linearization formula is L(x) = f(a) + f'(a)*(x-a) when, a = -1 f(a) = (-1)^4 + 3(-1)^2 = 4 f'(x) = 4*x^3 + 6*x f'(a) = 4(-1)^3 + 6(-1) = -10 so, L(x) = (4) + (-10)*(x-(-1)) = 4 - 10(x+1)
What is x to the 5th power divided by x to the 9th power?
0.5556
What is the derivative of 4 over x?
4/x can be written as 4x-1 (the power of negative 1 means it is the denominator of the fraction) 4*-1 = -4 Therefore, the derivative is -4x-2
How do you find the derivative of a function that has been raised to a power greater than 1?
Let us say that f(x)=x^4A derivative is the opposite to an integral.If you were to integrate x^4, the first process is taking the power [which in this case is 4], multiplying it by any value before the x [which is 1], then subtracting 1 from the initial power [4]. This leaves 4x^3. The final step is taking the integral of what is 'inside' the power [which is (x)], and multiplying this to the entire answer, which results in 4x^3 x 1 = 4x^3If you were to derive (x)^4, you would just add 1 to the power [4] to become (x)^5 then put the value of the power as the denominator and the function as a numerator. This leaves [(x^5)/(5)]To assure that the derivative is correct, integrate it. (x^5) would become 5x^4. Since (x^5) is over (5), [(5x^4)/(5)] cancels the 5 on the numerator and denominator, thus leaving the original function of x^4
What is the second derivative of square root x?
It is negative one divided by 4 multiplied by x to the power of 1.5 -1/(4(x^1.5))
