# How do you use the definition of a derivative to find the derivative of #-(2/3x) #?

We have:

By signing up, you agree to our Terms of Service and Privacy Policy

To find the derivative of -(2/3x) using the definition of a derivative, we apply the definition:

[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}]

For the function (f(x) = -(2/3x)), we substitute into the definition:

[f'(x) = \lim_{h \to 0} \frac{-(2/3(x+h)) - (-(2/3x))}{h}]

Simplify the expression:

[f'(x) = \lim_{h \to 0} \frac{-2/3(x+h) + 2/3x}{h}] [f'(x) = \lim_{h \to 0} \frac{-2x/3 - 2h/3 + 2x/3}{h}] [f'(x) = \lim_{h \to 0} \frac{-2h/3}{h}] [f'(x) = \lim_{h \to 0} \frac{-2/3}{1}] [f'(x) = -\frac{2}{3}]

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How do you find f'(x) using the definition of a derivative for #f(x)=1/sqrt(x)#?
- Using the limit definition, how do you differentiate #f(x)=1-x^2#?
- How do you find the points where the graph of the function #f(x) = x^4-4x+5# has horizontal tangents and what is the equation?
- What is the equation of the tangent to the curve # y=9tanx # at the point where #x=(2pi)/3#?
- What is the equation of the normal line of #f(x)=x^2-x+5# at #x=2#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7