please someone i know i can do them seperate using a rule, but its hard this one
Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx((x^(1/3) + sqrt(x) + x)/x^2)
The derivative of y is y'(x):
y'(x) = d/dx((x^(1/3) + sqrt(x) + x)/x^2)
Use the product rule, d/dx(u v) = v ( du)/( dx) + u ( dv)/( dx), where u = 1/x^2 and v = x + sqrt(x) + x^(1/3):
y'(x) = (x^(1/3) + sqrt(x) + x) d/dx(1/x^2) + (d/dx(x^(1/3) + sqrt(x) + x))/(x^2)
Use the power rule, d/dx(x^n) = n x^(n - 1), where n = -2: d/dx(1/x^2) = d/dx(x^(-2)) = -2 x^(-3):
y'(x) = (d/dx(x^(1/3) + sqrt(x) + x))/x^2 + (x^(1/3) + sqrt(x) + x) (-2)/x^3
Differentiate the sum term by term:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (d/dx(x^(1/3)) + d/dx(sqrt(x)) + d/dx(x))/x^2
Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1/3: d/dx(x^(1/3)) = d/dx(x^(1/3)) = x^(-2/3)/3:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (d/dx(sqrt(x)) + d/dx(x) + 1/(3 x^(2/3)))/x^2
Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 1/2: d/dx(sqrt(x)) = d/dx(x^(1/2)) = x^(-1/2)/2:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (1/(3 x^(2/3)) + d/dx(x) + 1/(2 sqrt(x)))/x^2
The derivative of x is 1:
y'(x) = -(2 (x^(1/3) + sqrt(x) + x))/x^3 + (1/(3 x^(2/3)) + 1/(2 sqrt(x)) + 1)/x^2
Simplify the expression:
y'(x) = (1 + 1/(3 x^(2/3)) + 1/(2 sqrt(x)))/x^2 - (2 (x^(1/3) + sqrt(x) + x))/x^3 - Courtesy of "Mathmatica 11 Home Edition"
