Implicit Differentiation: Example 2

00:01 So once you've attempted all of these questions in implicit differentiation, let's now go through them to check how your solutions went.

00:09 The first question reads 2 x plus y to the 5 equals to 3.

00:15 Okay, first of all, you can rearrange this equation to get y by itself, because it's not that complicated, but since we're practicing implicit differentiation, since it's supposed to make our lives easier let's just do this implicitly.

00:35 We're going to differentiate each individual term, but we're going to do it either with respect to x or with respect to y.

00:42 2x with respect to x just differentiates to 2, remember we are now differentiating a y function which we just do as you usually would differentiate x, so y would differentiate to 5y to the power of 4, bring the power down, decreased the power by one, but remember, and this is really important that every time you differentiate y, you must multiply it with the factor of dy/dx.

01:07 3 differentiates to zero and please be aware that all constants disappear, this is quite a common mistake that people make that they leave those constant there, because we're so focused on differentiating the y's.

01:19 We just rearrange this equation, we'll leave the dy/dx here, let's just leave 5y to the 4, dy/dx here, move your 2 to the other side and then finally dy/dx is just going to be minus 2 over 5y to the power of 4.

01:37 And again, you can substitute numbers in, to find the gradient at specific points.