# Why does differentiation find the approximate value?

1. Sep 21, 2014

### Aichuk

Like imagine I have to find the cube root of 8.03. So I cube 8 and use the dy=8+dy/dx*(8.03-8.00) formula. But why is this finding the value of cube root of 8.03 and why is this value approximate instead of exact?

2. Sep 21, 2014

### Number Nine

What is the derivative at 8, conceptually?

3. Sep 21, 2014

### Aichuk

1/12

4. Sep 21, 2014

### Staff: Mentor

Look at the graph of $y = x^{1/3}$. The derivative, dy/dx, gives the slope of the tangent line to this curve. The formula you show gives you the y values on the tangent line, which is close to, but slightly different from the y values on the curve.

Since the tangent line at (8, 2) lies above the curve $y = x^{1/3}$ , the approximate values will be a little larger than the values on the curve.

5. Sep 21, 2014

### Staff: Mentor

The value of the derivative isn't what Number Nine was asking. He was asking about the meaning of the derivative at that point.

6. Sep 22, 2014

### Aichuk

Your replies made me realize all the holes in my calculus knowledge. I've been taught calculus in school just through formulas so it looks like my concepts are very weak. I have no idea what to do about it because my textbook also only uses formulas.

7. Sep 22, 2014

### Staff: Mentor

Most calculus texts contain formulas, but they usually contain explanatory text as well. Are you saying that your textbook doesn't have explanations to go with the formulas? Sometimes students focus on the formulas and ignore the surrounding text.

8. Sep 22, 2014

### HakimPhilo

Basically it comes from this really simple idea: "The tangent line to $x_0$ resembles the curve near $x_0$". For instance the curve defined by $y=\sin x$ resembles the tangent line to it at $0$ near zero:

For example it is really hard to determine $\sin(0.1\,\rm rad)$ without using a calculator. But since the tangent line has a simple form, namely $y=mx+b$, one can easily exploit the fact that the tangent line resembles the curve of $\sin x$ near $0$ since $0.1$ is approximately zero to find a rough estimate for $\sin(0.1\,\rm rad)$.

9. Sep 24, 2014

### Aichuk

My textbook contains examples of HOW to use, but not WHY

10. Sep 24, 2014

### Staff: Mentor

Assuming each section of your textbook ends with a set of problems, maybe these problems are the WHY the formulas are used.