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### Course: APยฎ๏ธ/College Calculus BCย >ย Unit 3

Lesson 6: Selecting procedures for calculating derivatives: strategy

# Manipulating functions before differentiation

Sometimes, before differentiating a function, we can rewrite it so the process of differentiation is faster and easier.

## Want to join the conversation?

• At , isn't the equation not differentiable at x = 1? Do you need to make a note of "for x =/= 1" when defining the derivative in this way?
• Good spot! You are correct, Sal should have specified that the result is true for `x โ  1` .

I've submitted a "clarification".
• in the second last example, why couldn't you simply use the power rule, instead of the power rule AND the chain rule? Thanks :)
• When the function consists of a function inside a function that cannot be simplified algebraically, then the chain rule must be used.
• I've got a slightly unrelated question here, how does one factorize that quickly? When I see `x^2 + x - 2` I solve it like this:
``x^2 - x + 2x - 2x(x - 1) + 2(x - 1)(x + 2)(x - 1)``

I have no clue how one instantly does it...I guess knowing that I convert x into 2x - x I should probably get 2 and -1 but I think there will be cases for which it won't be true? So what's the consistent way of doing it instantly while being sure that it works?
• When the squared term has no coesfficient, it's quite easy: find two numbers that multiplied give you the constant and that when added give you the coefficient of x. Then arrange them into (x+a)(x+b). Be sure to keep the correct coefficients. In yours, the numbers that multiplied are -2 are 1, -2 and -1, 2. The sums of these are -1 and 1, respectively. The answer easily found to be -1 and 2, for the same answer as you got. Hope that helps!
• At , x^-2*x^1/2 is x^-1/2 not x^-3/2. Not a major mistake, but just thought I'd let you know
• The video is correct.

x^(-2) * x^(1/2)
= x^(-2 + 1/2)
= x^(-4/2 + 1/2)
= x^(-3/2)
• are there any case where the qoutient rule is better or is it completely useless?
(1 vote)
• You don't need to use the quotient rule, but it is usually easier than using a combination of the product, chain, and power rules.
• For the last example (at ) why doesn't the power rule work? When I tried it on my own, I brought the 2 to the front and changed the exponent to 1, making it 2(2x+1)^1. This gives 4x+2, which is the wrong answer.
• Here the expression (2x+1)^2 comprises of a function inside another function. That is: f(x)= 2x+1 and g(x)= x^2, so g(f(x))= (2x+1)^2.
So, here the chain rule is applied by first differentiating the outside function g(x) using the power rule which equals 2(2x+1)^1, which is also what you have done. This is then multipled by the derivative of the inside function f(x) that is 2x+1 which is 2.
So, the derivative of (2x+1)^2 = 2(2x+1)*2 = 4(2x+1) = 8x+4 as shown in
(1 vote)
• I got stuck at the first point if anyone can help me I will be thankful, in the y=x^2-x-2/x-1, and after we simplified it gives us d/dx[X+2]* it means if we take the derivative of it it will give us [1 ] it means for any x equal any number it doesn't matter because the result will be ONE for example if we take *x=6 for y=x^2-x-2/x-1 the slope of the tangent line will be ONE. if x=2,3,4,5,6 the slope of the tangent line will be ONE according to instructor cuz d/dx[X+2] =1.
and using the quotient rule I got *dy/dx=[x^2-x-2/x-1] = X^2-2X+3/(X-1)^2 * this seems more reasonable to me cuz I can now substitute X with any value and it will give me the slope of the tangent line at that point so why that difference
I know limits using limit after simplified it will be x+2 for X not equal to 1 cuz if the x is equal to one in the denominator it will give us vertical asymptote hence there is no limit at this point thus there no derivative at that point
(1 vote)
• Your question is not well written. You need to use brackets remember BODMAS.

Your calculation using quotient rule seems to be incorrect. As a double check use differentiation calculator.

By the way there seems to be typo in your question. Remember how to factor quadratics.
(1 vote)
• It seems as though the 2nd example at is incorrect.
I keep getting x^2+5/x^2

Am I missing something or did Sal slip up?
(1 vote)
• You are correct, but you miss the last step.
Quotient rule =(f'(x)g(x)-f(x)g'(x))/g(x)^2
Lets define every term
f(x)=x^2+2x-5
f'(x)=2x+2
g(x)=x
g'(x)=1
Now plug everything back in the formula
=(2x+2)*x-(x^2+2x-5)*1)/x^2
=(2x^2+2x-x^2-2x+5)/x^2
=(x^2+5)/x^2
Which equals 1+5x^(-2) -- like the video at
Sal's way is preferred in my opinion
(1 vote)
• Can I do a long division before differentiation to simplify functions? For example, in case if I have...
(x^3-4x^2+2x+8)/(x2-3x)
something like that...
• It really depends on the function. Long division can potentially change a function. For example, it can't be done here. Let me explain with a simpler example.

Let's say we have the function f(x) = (x^2 - 4x)/(x-4). Now, this can be simplified by long division by canceling x-4. However, this is a new function g(x) = x. The reason I say it is a new function is that f(4) does not exist, but g(4) does exist. This makes them different functions, with potentially different derivatives.

Of course, this is not always the case. If the denominator is a function like h(x)= x^2 + x + 1 which has no real factors, then yes, cancellation is allowed because, h(x) will never equal zero, (in fact, h(x) is greater than 3/4 for all real x) and so, if it is canceled out (by a similar term in the numerator), then the function's integrity is maintained.

To sum up, long division can be applied if the denominator does not equal zero for any real number. Therefore, it is inapplicable here.