Use the shortcut rules to mentally calculate the derivative of the given function
f(x) = −x + (3/x) + 1
After deriving it, I got
-1+(0/1)+0= -1
But this answer is wrong, and I don't know why.
Thank You
The derivative of f(x) is
-1-3/x²
Don't know how you got -1, or what the question really is.
How did you get the answer? I don't quite understand. Thanks
If f(x)=g(x)+h(x), then
linearity gives
f'(x)=g'(x)+h'(x)
so
f(x)=-x + (3/x) + 1
f'(x)= (-x)' + (3/x)' + (1)'
=-1 -3/x² + 0
=-1 -3/x²
The second term is obtained by the power rule:
d(1/x)/dx = -1/x²
To calculate the derivative of the given function, f(x) = -x + (3/x) + 1, using the shortcut rules, we need to apply the Power Rule, Quotient Rule, and Constant Rule.
1. Power Rule: The derivative of x^n with respect to x is n * x^(n-1).
In the given function, the power rule applies to the term (3/x). So, the derivative of (3/x) with respect to x is:
(3/x)' = 3 * (x^(-1-1)) = 3 * x^(-2) = 3/x^2
2. Quotient Rule: The derivative of f(x)/g(x) with respect to x is [g(x) * f'(x) - f(x) * g'(x)] / [g(x)]^2.
In the given function, we have -x as f(x) and 3/x as g(x). Applying the quotient rule, we get:
[f(x)/g(x)]' = [(-x)' * (3/x) - (-x) * (3/x^2)] / [(3/x)]^2
= [(-1) * (3/x) - (-x) * (3/x^2)] / [(3/x)]^2
= [-3/x + (3x/x^2)] / [(3/x)^2]
= [-3/x + 3x/x^2] / [9/x^2]
= (-3x + 3x^2) / 9x^2
3. Constant Rule: The derivative of a constant is always zero.
In the given function, the constant term is 1. So, the derivative of 1 with respect to x is zero.
Now, we sum up the derivatives of all the terms in the given function:
f'(x) = (-x)' + (3/x)' + (1)'
= -1 + (3/x^2) + 0
= -1 + 3/x^2
Your answer, -1, doesn't match the correct derivative. The derivative should be -1 + 3/x^2.
Note: While differentiating, be careful with the computations and make sure not to confuse addition and subtraction signs or make any arithmetic mistakes. Remember to apply the derivative rules correctly to each term.