Derivada de una función potencial Derivar funciones potenciales Partiendo de la derivada de la función identidad que, por la misma definición de derivada, es la unidad, vamos incrementando el exponente de uno en uno para los primeros valores de n n n , aplicando la derivada de un producto.
( x ) ′ = 1 ( x 2 ) ′ = ( x ⋅ x ) ′ = ( x ) ′ ⋅ x + x ⋅ ( x ) ′ = 2 x = 2 x 2 − 1 ( x 3 ) ′ = ( x ⋅ x 2 ) ′ = ( x ) ′ ⋅ x 2 + x ⋅ ( x 2 ) ′ = x 2 + x ⋅ 2 x = 3 x 2 = 3 x 3 − 1 ( x 4 ) ′ = ( x ⋅ x 3 ) ′ = ( x ) ′ ⋅ x 3 + x ⋅ ( x 3 ) ′ = x 3 + x ⋅ 3 x 2 = 4 x 3 = 4 x 4 − 1 . . . ( x n ) ′ = n x n − 1 (x)'=1\\(x^2)'=(x\cdot x)'=(x)'\cdot x+x\cdot (x)'=2x=2x^{2-1}\\(x^3)'=(x\cdot x^2)'=(x)'\cdot x^2+x\cdot (x^2)'=x^2+x\cdot2x=3x^{2}=3x^{3-1}\\(x^4)'=(x\cdot x^3)'=(x)'\cdot x^3+x\cdot (x^3)'=x^3+x\cdot3x^2=4x^{3}=4x^{4-1}\\...\\(x^n)'=nx^{n-1} ( x ) ′ = 1 ( x 2 ) ′ = ( x ⋅ x ) ′ = ( x ) ′ ⋅ x + x ⋅ ( x ) ′ = 2 x = 2 x 2 − 1 ( x 3 ) ′ = ( x ⋅ x 2 ) ′ = ( x ) ′ ⋅ x 2 + x ⋅ ( x 2 ) ′ = x 2 + x ⋅ 2 x = 3 x 2 = 3 x 3 − 1 ( x 4 ) ′ = ( x ⋅ x 3 ) ′ = ( x ) ′ ⋅ x 3 + x ⋅ ( x 3 ) ′ = x 3 + x ⋅ 3 x 2 = 4 x 3 = 4 x 4 − 1 ... ( x n ) ′ = n x n − 1
Función derivada para funciones potenciales
f ( x ) = x n ⟹ f ′ ( x ) = n x n − 1 f(x)=x^n\implies f'(x)=nx^{n-1} f ( x ) = x n ⟹ f ′ ( x ) = n x n − 1
Ejemplo Calcula la derivada de f ( x ) = x 3 5 {\it f(x)=\sqrt[5]{x^3}} f ( x ) = 5 x 3
En primer lugar, reescribimos la raíz como potencia de exponente fraccionario.
x 3 5 = x 3 5 {\sqrt[5]{x^3}=x^{\frac{3}{5}}} 5 x 3 = x 5 3
Después, aplicamos la regla para derivar una función potencial
g ( x ) = x a ⟹ g ′ ( x ) = a x a − 1 f ( x ) = x 3 5 ⟹ f ′ ′ ( x ) = 3 5 ⋅ x − 2 5 = 3 5 x 2 5 { g(x)=x^a\implies g'(x)=ax^{a-1}}\\{ f(x)=x^{\frac{3}{5}}\implies f\ ''(x)=\cfrac{3}{5}\cdot x^{\frac{-2}{5}}=\cfrac{3}{5\sqrt[5]{x^2}}} g ( x ) = x a ⟹ g ′ ( x ) = a x a − 1 f ( x ) = x 5 3 ⟹ f ′′ ( x ) = 5 3 ⋅ x 5 − 2 = 5 5 x 2 3