# Simple Differential Problem (1 Viewer)

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#### JoshHolloway

This is a problem from a Calc 1 text book, I just can't figure out where they get dx from. The question is:
Use differentials to approximate the value of the expresesion.
$$\sqrt{99.4}$$

the answer in the solution manual says:
Let $$f(x)=\sqrt{x}, x=100, dx=-.6$$

then it solves the problem. But what I don't understand is where the heck is this dx quantity derived from?

#### HallsofIvy

$$f(x)=\sqrt{x}, x=100, dx=-.6$$

Your textbook is using the fact that df= f'(x)dx so that f(x+dx) is approximately
f(x)+ df= f'(x)dx.
Here, you want to evaluate f(99.4) and it easy to see that f(100)= 10 so take x= 100 and x+ dx= 99.4. What is dx?

#### JoshHolloway

Wow, I get it. It's so simple. Dx is -0.6. This also makes it more clear why it is called a differential. Thanks Ivy! Hey Ivy, are these differentials the same thing that differential equations are based on?

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#### mustafa

JoshHolloway said:
Wow, I get it. It's so simple. Dx is -0.6. This also makes it more clear why it is called a differential. Thanks Ivy! Hey Ivy, are these differentials the same thing that differential equations are based on?
Though similar, they are not exactly the same thing.

In terms of differentials df=f'(x)dx is true but taking dx= -0.6 is only an approximation. In the true sense a differential can be considered as an infinitesimally small change in the variable.

You might be knowing the Taylor's series expansion of f(x+h)

f(x+h) = f(x) + hf'(x) + (h^2)/2! f"(x) + (h^3)/3! f"'(x) + ...

If h is very small relative to x, the first order approximation can be obtained as f(x+h) = f(x) + hf'(x), which is the same as the equation with differentials
f(x+dx) = f(x) + f'(x)dx

What I want to imply is that the dx in your question is an approximation but differentials in a differential equation do not signify any approximations.

#### HallsofIvy

Strictly speaking, differentials, like "dx", are "infinitesmals". When you write something like "dx= -0.6", it really is $$\Delta x$$, meaning a small change in x. One way of defining the derivative is $$lim_{\Delta x->0} \frac{\Delta y}{\Delta x}$$. You can then use $$\Delta x$$ to approximate dx:
$$\frac{\Delta y}{\Delta x}$$ is approximately $$\frac{dy}{dx}$$ so $$\Delta y$$ is approximately $$$$\frac{dy}{dx}$$\Delta x$$. The smaller $$\Delta x$$ is, the better the approximation.

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