00:01
Hi, there is a question which says that simplify the differential quotient for y equal to x square, fx equal to 2x square, 2x square and sketch three tangents on the graph at a point when x is minus 2, 0 and 1.
00:20
I need to first simplify fx plus h minus fx.
00:27
Okay, so fx plus h minus fx by h will be 2 into x plus h whole square minus 2x square by h, 2 will be taken as common, so x plus h whole square minus x square divided by h.
00:49
So 2, this is a square minus b square, a plus b into a minus b divided by h.
01:00
This will get cancelled out.
01:02
So 2 into 2x plus h into h by h, h gets cancelled out, 2 into 2x plus h.
01:16
Second, f prime x will be when limit will h approach to 0.
01:25
So 2x plus h, when we plug in h equal to 0, this will become 4x.
01:32
4x, so to find slope of tangent at x equal to minus 2, that means f prime minus 2 is 4 into minus 2, which is minus 8.
01:44
So equation of the tangent at x equal to minus 2, when x equal to minus 2, the point will become minus 2 comma 8.
01:52
This is the point of tangency by plugging in y equal to 2x square.
02:02
And the slope of the equation of tangent will be y minus y1, y minus y1 equal to slope x minus x1.
02:15
So y equal to minus 8x minus 16 plus 8, y equal to minus 8x minus 8, second is when x equal to minus 1...