If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Differential Calculus

Course: Differential Calculus>Unit 3

Lesson 7: Strategy in differentiating functions

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 ≠ 1` .

I've submitted a "clarification".
• 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)
• 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!
• 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)
• are there any case where the qoutient rule is better or is it completely useless?
• 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.
• 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)
• For the last example you could just write (2x +1)(2x + 1) then have the realization that the applying product rule would simplify the expression to 2(f'(x)(f(x)). Ahhh the beauty of mathematics.
• 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...