Find the derivative of [(2+x)/(x-3)]^(2/5)
I tried the power of a function rule, quotient rule, the chain rule but keep getting stuck.
Yes, you have to use the chain rule here.
Step one:
Take [(2+x)/(x-3)] as y.
d(y^2/5)/dx = (2/5)y^(-3/5), as per the exponent rule for differentiation.
Next, apply the quotient rule on [(2+x)/(x-3)] to get its derivative.
Multiply the answer obtained in both steps for the final answer.
I got (2/5)[(2+x/x-3)^(-3/5)] (-5/(x-3)^2)
Now I am having troubles simplifying it.
To find the derivative of the function [(2+x)/(x-3)]^(2/5), we can use a combination of the quotient rule, the chain rule, and the power rule. Here is the step-by-step process:
Step 1: Rewrite the function using exponent rules.
[(2+x)/(x-3)]^(2/5) = ((2+x)^(2/5))/((x-3)^(2/5))
Step 2: Apply the quotient rule to the function.
Let f(x) = (2+x)^(2/5) and g(x) = (x-3)^(2/5).
Using the quotient rule, the derivative of the function is given by:
[f'(x)g(x) - g'(x)f(x)] / [g(x)]^2
Step 3: Find the derivatives of f(x) and g(x).
Applying the chain rule to f(x) and g(x) individually:
f'(x) = (2/5) * (2+x)^(-3/5) * (d/dx)(2+x)
g'(x) = (2/5) * (x-3)^(-3/5) * (d/dx)(x-3)
Step 4: Calculate the derivatives (d/dx)(2+x) and (d/dx)(x-3).
(d/dx)(2+x) = 1 (because the derivative of a constant + x is simply 1)
(d/dx)(x-3) = 1 (the same reasoning as above)
Step 5: Plug the derivatives obtained in Step 4 into the derivatives of f(x) and g(x).
f'(x) = (2/5) * (2+x)^(-3/5) * 1 = (2/5)(2+x)^(-3/5)
g'(x) = (2/5) * (x-3)^(-3/5) * 1 = (2/5)(x-3)^(-3/5)
Step 6: Substitute f'(x), g'(x), f(x), and g(x) into the quotient rule formula from Step 2.
[f'(x)g(x) - g'(x)f(x)] / [g(x)]^2 = [(2/5)(2+x)^(-3/5) * (x-3)^(2/5) - (2/5)(x-3)^(-3/5) * (2+x)^(2/5)] / [(x-3)^(2/5)]^2
Thus, the derivative of [(2+x)/(x-3)]^(2/5) is [(2/5)(2+x)^(-3/5) * (x-3)^(2/5) - (2/5)(x-3)^(-3/5) * (2+x)^(2/5)] / [(x-3)^(4/5)].