taking derivatives: f_1 = (x^2 + x)^2, f_2 = -1/(x^2), f_3 = 1/x, f_4 = ...

[fluffy]

Hi,

I have to take derivatives but I'm somewhat stuck.

f₁=(x²+x)² | Solution: (x²+x)(4x+2) | inner derivation [2x+1] multiplied with outer derivation [2(x²+x)], why do I have to multiply the inner and not outer derivation with 2?
f₂=(-1)/x² | Solution: 2/x³ | When I got fractions to derive, can I always multiply by (-1), then add 1 to the exponent of the denominator, while multiplying the numerator with the exponent of denominator? I don't really get it.
f₃=1/x | Solution: (-1)/x² | Same as f₂. But the "rule" doesn't seem to apply. Is there some algebra magic behind that?
f₄=-(2h(x))/x | Solution: (-2xh'(x)+2h(x))/x² | Seeing such a term, how do I start?
f₅=4^{1-x^2} | Solution: (4^1-x2)(-2x*ln(4)) | My approach would be: (4(1-x²))^{-x^2}, so why's there the logarithm?
f₆=2(x²+x)(2x+1) | Solution: 12x²+12x+2 | What is the approach here? Inner multiplied with outer derivations? What about the coefficient "2"?

How can I write fractions? Also, what do I actually do when I take a derivative? E.g. x², the parabola, the derivative is 2x, a line which grows gradually at a bit more than 45° angle, intersecting x at (0, 0), positive for x>0, negative for x<0. What does that actually mean? I don't see the point on why we take derivatives. For optimization tasks we always set the first order derivative to 0. Why is that?

Thanks for your help!

 

[dblu]

My thoughts:


f1: Nothing says you have to multiply the inner derivation by 2, the author of the solution simpy chose to do so. Leaving the answer as 2(x^2+x)(2x+1), multiplying by outer derivation (2x^2+2x)(2x+1), or multiplying all terms, 4x^3+6x^2+2x, are all equally valid and equivalent answers.


f2: The way to approach this problem is to recognize that what you are deriving is -(x^-2). Applying the power rule (nx^n-1), it becomes 2(x^-3), which can be written as 2/x^3. I would suggest you become comfortable with converting negative exponents into fractions and vice versa.


f3: Same as f2. Recognize that what you are deriving is x^-1, apply the power rule as before, then convert back into a fraction (to eliminate the negative exponent).


f4: Don't let the undefined function h(x) throw you off. It is a term just like any other, just remember that you have to use the chain rule on it, and since it isn't defined, the best you can do is leave it as h(x) (or h'(x) to indicate when you've taken the derivative of it) in your solution.


f5: Remember that a^x is the inverse of the function f(x)=log_bx. So if y=b^x, we can rewrite as x = log_by and use implicit differentiation to obtain 1 = (1/y*ln(b))*dy/dx. Solving for dy/dx and substituting b^x for y, we get dy/dx=y ln(b) = b^x ln b. So we see that the derivative of b^x is b^x ln(b). This is where the ln comes from. Try using this along with the chain rule and you should arrive at the correct solution.


f6: I would get rid of the coefficient before taking the derivative by distributing it into one of the binomials. Then apply product rule, foil the binomials, and combine like terms. Manipulating the function prior to taking the derivative to make the process easier is absolutely legal (and in some cases, necessary).

I'm not sure what you mean by how can you write fractions. As far as what are you doing when you take the derivative and the point of it, I'll let somebody else take a stab at it, as my explanation will probably not make anything clearer to you.

Hope this helps!

 

[ksdhart]

I'll tackle your questions at the bottom first, then come back to the actual problems in a moment.

How can I write fractions?

If you're asking how you can write math symbols on this forum, that's called LaTeX. It can seem daunting at first to learn, but it's actually fairly straightforward. If you wanted to post the following problem:

\(\displaystyle \displaystyle \frac{d}{dx}\left(\frac{x^2+2}{x^3+7}\right)\)

You'd type this in your message:

[tex]\frac{d}{dx} ( \frac{x^2+2}{x^3+7} )[/tex]

I find this PDF to be useful. It has a large list of math symbols and their LaTeX equivalent.

Also, what do I actually do when I take a derivative?

The derivative represents the slope of a function. When you take the derivative of a function, you get another function, into which you can plug a specific value of your variable to get the slope at that point. Well, more specifically, the slope of the line tangent to that point, because curves don't really have slopes, as such.

For optimization tasks we always set the first order derivative to 0. Why is that?

When you're working an optimization problem, you're trying to find the maximum or minimum of a function, possibly given some constraints. If you think of that in terms of the graph, you'll see that anywhere there's a maximum or a minimum, the slope changes signs at that point. And since one of the requirements for a function to be differentiable at a point is to be continuous at that point, the slope must pass through 0 in order to change signs. Therefore, we can say that the only possible points where a function can have a maximum or a minimum is where the slope (aka first derivative) is 0.

Now, as for your problems. Most of them, with the exception of #4 and #5 are examples of one of the rules for taking derivatives. Your textbook should have these rules somewhere, though you might have to hunt for them and look a bit ahead of where you are in your class at the moment to find them.

The first one is what's called the chain rule. Your derivative looks like this:

\(\displaystyle \frac{d}{dx}\left(\left[g\left(x\right)\right]^2\right)\), where \(\displaystyle g\left(x\right)=x^2+x\)

The chain rule tells you to take the derivative of the outer function, so the derivative of g(x)^2 is 2g(x). Then you multiply by the derivative of the inner function. And the derivative of x^2+x is 2x+1. Putting it all together, we have the given answer of 2(x^2+x)(2x+1) or (x^2+x)(4x+2).

The second and third problems are examples of the power rule. For any constant n:

\(\displaystyle \frac{d}{dx}\left(x^n\right)=n\cdot x^{n-1}\)

Also recall that you can write 1/x as x^-1 and hopefully you can see how to tackle problems 2 and 3.

Problems #4 and #5 are more tricky and explaining them will take more time than I have right at this moment. If no one else has explained them, I will tackle them when I get a chance.

Problem #6 is an example of the product rule. Given two functions f(x) and g(x):

\(\displaystyle \frac{d}{dx}\left(f\left(x\right)\cdot g\left(x\right)\right)=\frac{df}{dx}\cdot g\left(x\right)+\frac{dg}{dx}\cdot f\left(x\right)\)

To put that into words, you might say "Derivative of the first function, times the second. Plus the derivative of the second function times the first." In your example, begin by pulling out the constant out front of the derivative and then let:

\(\displaystyle f\left(x\right)=x^2+x\) and \(\displaystyle g\left(x\right)=2x+1\)

What do you get when you apply the product rule?

I hope these explanations were of help to you.