# Approximation by differential

1. Jun 16, 2017

### Karol

1. The problem statement, all variables and given/known data
Approximate $~\sqrt[4]{17}~$ by use of differential

2. Relevant equations
Differential: $~dy=f(x)~dx$

3. The attempt at a solution
$$y=\sqrt[4]{x},~~dy=\frac{1}{4}x^{-3/4}=\frac{1}{4\sqrt[4]{x^3}}$$
$$\sqrt[4]{16}=2,~~dx=1,~~dy=\frac{1}{4\sqrt[4]{x^2}}\cdot 1=0.149$$
$$\sqrt[4]{17}=2.031$$
The error is too big

2. Jun 17, 2017

### Ray Vickson

Nevertheless, your result is correct.

If you want better accuracy you need to take additional, higher order terms (2nd derivatives, maybe 3rd derivatives,etc).

3. Jun 17, 2017

### Karol

Thank you Ray

4. Jun 17, 2017

### Staff: Mentor

Here is how I would write things.
$f(x + \Delta x) \approx f(x) + df \approx f(x) + f'(x) \Delta x = x^{1/4} + \frac 1 4 x^{-3/4} \Delta x$
When $x = 16$ and $\Delta x = 1$, we have
$\sqrt[4]{16 + 1} \approx 16^{1/4} + \frac 1 4 \frac 1 {16^{3/4}} \cdot 1 = 2 + \frac 1 32 = 2.03125$
By calculator, $\sqrt[4]{17} \approx 2.03054$. Since $\Delta x = 1$ is relatively large in comparison to x = 17, the approximation using differentials isn't all that accurate. If $\Delta x$ were smaller, the approximation would be better.

5. Jun 17, 2017

### Karol

Thank you Mark44
By the way, how do i copy your names here, i write them again. when i pause the mouse on your name it becomes a pointer and there is no option to copy

6. Jun 17, 2017

### WWGD

Just hit the 'Reply' button on the lower right of the post. You may also hite the 'Quote' button , also on the lower right.
BTW, small mistake: dy=f'(x)dx , not dy=f(x)dx , unless f(x)=f'(x).

7. Jun 18, 2017

### Karol

Thanks, but i mean i want to copy your name, WWGD, to here, instead of looking and typing it. i usually thank every one that answered my question

8. Jun 20, 2017

### WWGD

Maybe you can just use the 'Like' button as a means of thanking.