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.

Main content

# Differentiating functions: Find the error

AP.CALC:
FUN‑3 (EU)

## Video transcript

we're going to do in this video is look at the work of other people as they try to take derivatives and see if their reasoning is correct and if it's not correct try to identify what they should have done or where their reasoning went wrong so over here it says Nate tried to find the derivative of x squared plus 5x times sine of X here is his work is Nate's work correct if not what's his mistake so pause the video and see if you can answer this his Nate's work correct and if not what his mistake all right so I'm assuming you've had a go at it so let's work through this step by step so over here he's just trying to apply the derivative operator to the expression which is exactly what he needed to do he's trying to find the derivative of this thing and he says ok this is a product of two expressions and then he says okay well this is going to be the same thing as the derivative or this is the same thing as the product of the derivatives now this is a problem you are probably familiar if I take the derivative of the sum of two things so the derivative with respect to X of f of X plus G of X that indeed is equal to the derivative of the first F prime of X plus the derivative of the second but that is not true if we are dealing with the product of functions the derivative with respect to X of f of X G of X is not necessarily maybe there's some very special circumstances but in general it's not going to be just the product of the derivative it's not going to be just F prime of X G prime of X here we would want to apply the product rule this is going to be equal to this is going to be equal to the derivative of the first function times the second function plus the derivative well let me write it this way plus the first function not taking a straight if time's the derivative of the second function so he should have applied the product rule here and so let's do that just to see what his answer should have been so what he should have done here I'll get my correcting red pen out here say no that's not what he should have done he says let's take the derivative of this first thing so actually let me do the color coded so the derivative of this is 2x plus 5 so it should have been 2x plus 5 times the second thing so times sine of x times do it in another color times sine of X and then to that he would add the first thing which is x squared plus 5x times the derivative of the second thing so the derivative of sine of X is cosine of X so this is what he should what we should have been seeing at this step right over here he shouldn't have just taken the product of the derivatives he should have applied the product rule so his work is not correct and his mistake is that he didn't apply the product rule he just took he just assumed that the derivative of the products is the same thing as the product of the derivatives there's two more examples okay so let's see it says Katie tried to find the derivative of 2x squared minus 4 all of that to the third power here is her work is Katie's work correct if not what is her mistake so once again pause the video see if you can figure it out alright now let's inspect Katie's work so she's taking the derivative of this and let's see over here it looks like she's taking the derivative of the entire expression with respect to the inner expression and that is close to applying the chain rule properly but it's not applying the chain rule properly so her work is not correct and our mistake is she's not correctly applying the chain rule just as a review the chain rule says look if we're trying to take the derivative with respect to X of of F of G of X F of G of X that this is going to be equal to this is going to be equal to the derivative of the whole thing with respect to G of X so I could write that as f prime of G of X F prime of G of X times the derivative of the inner function with respect to X times G prime of X so over here then we could view our F function it's the thing that takes its input and takes it to the third power and so this right over here is f prime of G of X so this thing is the F prime of G of X but she forgot to multiply it by the derivative of the inner function with respect to X so she forgot to multiply this times the derivative of 2x squared minus 4 with respect to X which is going to be let's see the derivative of 2x squared power rule 2 times 2 is 4 so it's going to be 4 X to the first and then the derivative of negative 4 is just a 0 so it's going to be times 4x so that's what she needed to do in order for it to be correct so she had to have this x for X here times 4x so not correct she didn't up correctly apply the chain rule so let's do another one of these so here says the Jomon tried to find the derivative of sine of 7x squared plus 4x here's his work is no Joe mons work correct if not what is his mistake pause the video see if you can figure it out all right so it's the derivative of sine of this expression so you'd want to use the chain rule and find using the chain rule you want to find the derivative of the outside function with respect to the inside so the derivative of sine of something with respect to that something is going to be cosine of that something so that's right that's right and then you want to multiply that times the derivative of the inside with respect to X so the derivative 7x squared is 14x derivative of 4x is 4 so this is actually that step looks good but then the Jomon does something strange over here this is the cosine of 7x squared plus 4x and then that whole thing times 14 X plus 4 but they get confused we're just looking at these parentheses and this tends to happen sometimes it's actually one of these key errors that the folks at the College Board ap folks about that when dealing with these transcendental functions cosine sin to sine tangent natural log that are written like this and people see the parentheses and see in other parentheses their brain this is oh let me multiply these two parentheses two expressions in parentheses but that's not right because if we were to add parentheses this is what this is implying so you can't just take the 14 X + 4 and multiply it by this and assuming you're taking the cosine of the whole thing so this is where the Jomon makes the mistake the work is not correct and the mistake is trying to multiply these two expressions and taking the cosine of the whole thing let's do one more of these I find these strangely fun all right this one is involved Tom tried to find the derivative of the square root of x over X to the fourth here is his work is Tom's work correct if not what's his mistake pause the video and see if you can figure that out so it looks like he's trying to apply the quotient rule so applying the quotient rule you would in the numerator you would take the derivative of the first expression times the second expression and then minus the first expression times the derivative of the second expression all of that over the or I should say the derivative the first the numerator expression over the times the denominator expression minus the numerator expression times the derivative of the denominator expression all of that over the denominator expression squared so this looks correct actually it's correct application of the quotient rule it looks like tom is correctly simplifying so the derivative of X to the 1/2 is 1/2 X to the negative 1/2 so that looks right derivative of X to the fourth is 4x to the third so that looks right all of this looks algebraically right and let's see when you simplify this so it's the X to the negative 1/2 times X to the fourth is indeed x2 well that's going to be X to the OH this correlates so this simplifies to that which looks correct and that simplifies to that which looks correct we're just using exponent properties there and then divided everything by let's see oh then everything in terms of X - 3.5 so so we're going to have negative negative 3.5 x to 3.5 and then you use exponent property so it actually looks like he did everything correctly this is the right answer now so his work is correct he did not make any mistakes but I would I do have a bone to pick so speak with Tom because he didn't have to apply the quotient rule here he did all of this Harry calculus and algebra but there could have been a very simple simplification he could have made up here and this is a key thing to realize he could have said hey you know what this is the same thing it's a derivative with respect to X of of X to the one-half that's what the square root of x is times X to the negative fourth power that's what the X that's what 1 over X to the fourth is and so let me color code it so that is the same thing as that and that is the same thing as that and you wouldn't even have to use a product rule here you could simplify this even further this is the same thing as the derivative with respect to X of this we have the same base we can add the exponents working the products it's going to be X to the negative 3 point X to the negative 3.5 and so you can just use the power rule so this is going to be equal to bring the negative 3.5 out front negative 3.5 x to the and then we just decrement this one by one we subtract one from that negative 4.5 power so as you can see he could have gotten this answer much much much much much much quicker but he didn't make any mistakes there's a little bit of a judgment error just immediately to going forth with the quotient rule which gets quite hairy quite fast
AP® is a registered trademark of the College Board, which has not reviewed this resource.