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Implicit Differentiation: Exercise 4

by Batool Akmal
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    00:01 And our last example, this looks fairly fun because I can see that you have y?s, you have x?s.

    00:07 I can also see a product in the middle, so I can see y multiplied by sine 2x and if you look closely, sin to x is also a function of a function.

    00:16 But let?s not panic, write it down and we'll do this one step at a time.

    00:21 So, let?s look at each term so we have 3x to the 4 plus y, sine of 2x, equals to y squared plus 4x.

    00:32 Remember to give some attention to each individual term, this one seems okay it?s got x?s on it, this one we?re going to differentiate y with respect to y but don?t forget to multiply it with dy/dx.

    00:45 This one looks okay, this one here, there?s a few different things happening but we?ll deal with that shortly, so, it?s a product.

    00:52 Let?s just do the easier ones first, so I?ve got 3x to the 4 bring the power down, decrease the power by one so that gives me 12x to the 3 plus, so let?s make some space here for our product rule, we'll gonna do that at the end y squared gives me 2y multiplied by dy/dx, because it was a y differential and 4x differentiates to 4.

    01:17 So there's quite a few things that your mind is doing at the same time with switching between differentiating y, differentiating x.

    01:23 Okay, now to this term here, we are going to use the product rule, so for the product rule remember to leave one term as it is.

    01:31 So let?s just leave this as it is to start with, y now we?re trying to differentiate sine 2x.

    01:38 so, just say it doesn?t get messier, I?m just going to do this here on the side, sine 2x, dy/dx.

    01:45 Remember, I?m just differentiating this with respect to x, so sine goes to cos, the whole thing, so the angle it remains is 2x.

    01:57 So it's the chain rule, so the whole function firstly you differentiate, so if it sine of anything it goes to cos of the same thing, and remember to multiply it with the differential of the inside, which is just 2.

    02:09 So the differential of sine 2x with respect to x is 2 cos 2x.

    02:14 So I?m gonna write that here, so, that?s 2 cos 2x.

    02:17 I?m gonna put the angle and bracket so we don?t get confused and times it with other things.

    02:21 Put a plus in the middle, we're now going to differentiate y, so we're going to differentiate this term here.

    02:29 Remember what to do when you differentiate y, you differentiate it with respect to y so that just gives you one, but also to multiply it with a factor dy/dx, and the sine 2x stays as it is, so let's squeeze that in.

    02:44 So we?ve done the entire differential, we?ve done the product rule here, we just now need to tidy this up and rearrange for dy/dx.

    02:54 So 12x cubed, plus, I?m gonna write this in order 2y cos 2x.

    03:00 I can multiply this through with the plus because it?s not gonna change anything, so, that?s fine, we can take it out of the bracket plus dy/dx sine 2x, should have written that the other way around but it doesn?t matter, and then I have 2y dy/dx plus 4. You can see that you have 2dy/dx terms here, this one and this one, and then you have lots of different numbers, if we just move this minus four to the other side, sorry plus 4 to the other side.

    03:32 So that gives me 12x cubed plus 2y cos of 2x minus 4 and then all my dy/dx is moved to the other side.

    03:43 So I?ve got 2y, dy/dx and then this is my second dy/dx term, and I?m moving to the other side of the equation so, minus sine 2x, multiplied it by dy/dx.

    03:55 I have written it the right way, usually we write dy/dx at the end of the expression.

    03:59 Take a common factor of dy/dx and this expression here, so, 12x cubed plus 2y cos of 2x minus 4, dy/dx is my common factor leading me with 2y minus sine 2x and you know what to do next, we just take it to the other side and divide. So, our gradient is 12x cubed, plus 2y cos of 2x minus 4, that?s all been divided by 2y minus sine 2x equals to dy/dx.

    04:39 So lots of trigs in this and lots of algebra, the important thing obviously is practicing the product for all to this little thing here is important, it?s also important to keep it in brackets because sometimes the sine in the outside might be negative.

    04:56 So if it is negative the sine inside the product will change.

    05:00 And to remember to leave one term as it is differentiate the other and then the one that you didn?t differentiate, you then differentiate the second time and leave the first term as it is.

    05:10 So, we?re all done at practicing the exercise questions.

    05:14 I hope this has been helpful if you had any problem go through them again with the solutions and hopefully you'll reach the correct answer this time.


    About the Lecture

    The lecture Implicit Differentiation: Exercise 4 by Batool Akmal is from the course Implicit Differentiation.


    Author of lecture Implicit Differentiation: Exercise 4

     Batool Akmal

    Batool Akmal


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