Sometimes it is required to find a root derivative in tasks of the mathematical analysis. Depending on statements of the problem, derivative of the "square root" function (cubic) is directly or by transformation of "root" to a power function with a fractional exponent.
It is required to you
- - pencil;
- - paper.
Instruction
1. Before finding a root derivative, pay attention to other functions which are present at a solvable example. If in a task there are many radicands, then use the following rule of finding of a derivative of a square root: (kh)' = 1/2kh.
2. And apply a formula to finding of a derivative of a cubic root: ³ √ x)' = 1/3 ³ √ x)² where through³ √ x the cubic root is designated from x.
3. If in the example intended for differentiation the variable in fractional degrees meets, then transfer designation of a root to a power function with the corresponding indicator. For a square root it will be degree ½, and for a cubic root – ⅓: √ x = x ^ ½,³ √ x = x ^ ⅓ where the symbol ^ designates exponentiation.
4. For finding of a derivative power function in general and х^½, x^⅓, in particular, use the following rule: (x ^ n)' = N* x^ (n-1). For a root derivative from this ratio follows: (х^½)' = ½ x ^ (-½) and (x^⅓)' = ⅓ x ^ (-⅔).
5. Having differentiated all roots, attentively look at other parts of an example. If in the answer at you very bulky expression turned out, then for certain it can be simplified. The majority of school examples is so that the small number or compact expression as a result turned out.
6. In many tasks on finding of a derivative, roots (square and cubic) meet together with other functions. To find a root derivative in this case, apply the following rules: • a constant derivative (constant number, C) equals to zero: With' = 0; • the constant multiplier is taken out for the sign of a derivative: (k*f)' = k * (f)' (f is any function); • the derivative of the sum of several functions equals to the sum of derivatives: (f + g)' = (f)' + (g)'; • the derivative of performing two functions equals … is not present, not to the work of derivatives, and the following expression: (fg)' = (f) 'g + f (g)'; • the derivative private also equals not private derivatives, and is according to the following rule: (f/g)' = ((f) 'g – f(g)') / g².