00:01
Given equation which is given as 4x square plus y square minus 8x plus 4y plus 4 equals to 0.
00:14
Now we have to use implicit differentiation to take the derivative with respect to x on both the side.
00:21
How we can do it? dy of dx 4x square plus y square minus 8x plus 4y plus 4 equals to dy by dx 0.
00:40
That will become 8x plus 2y minus 2y then there will be dy by dx.
00:52
That will be dy because we are differentiating with respect to x minus 8 plus 4 into dy by dx plus 0 and equals to 0.
01:06
Now again we can take the two common.
01:11
We will get x plus y into dy by dx minus 4 plus 2 into dy by dx equals to 0.
01:23
From there we can send x minus 4 in right side and dy by dx we can take common.
01:31
We will get y plus of 2 that is equals to 4 minus of that will be x that will 4x sorry so 4 minus of 4 of x.
01:45
Then again we can write it dy by dx equals to y plus of 2 divided by 4 minus of that will become sorry that will become 4 minus of 4x divided by 5 plus of 2.
02:03
Okay so finally we are we are having dy by dx equals to 4 minus of 4x divided by y plus 2.
02:12
Now what we will do? we will find out the original tangent on the curve is dy by dx.
02:19
In horizontal tangent on the curve dy by dx equals to 0 that is we can write is we can put 4 minus of 4x equals to 0...