Gradient of a curve at a point

[@cat_master4346]

can i just say this was really helpful, straight forward and detailed at the same time, this really helped with understanding, thank you very much, ive been struggling with this so much, so much that i have failed my test, but this really helps me understand, we need more teachers that can actually teach, my one just assigns work and expects us to know

[@zattack5757]

Really useful and straight to the point thanks

[@williamwade2674]

doing gods work

[@orotiktok9323]

Future savier from exam thanks so muck madam

[@habibazein716]

You saved my future ❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️❤️

[@KaieataTeambo]

How if fing the gradient of the tangent to the curve y=x^2 at the point (3,9)

[@88lbeal]

@KaieataTeambo at GCSE to find the gradient of a curve at a point you draw a tangent at that point because the gradients will be 'equivalent'. Then you learn to be more accurate you differentiate. So for your equation dy/dx = 2x now the point you gave was (3,9) so the x value of the coordinate is 3 so sub this in to dy/dx. 2x = 2 x 3 = 6

[@bhuvanareddy6809]

Mam can u give us theorem proof for gradient of curve is dy/dx?

[@Gabriel-zd8iy]

What about the 8 in (2,8)?

[@yuanxinling6430]

sub in 2

[@88lbeal]

The easiest way to answer this is the equation only has the x variable in it so you only use the x value. The 8 is the 'output' for the equation. That has no use to us now as we are not using the equation we are using the differentiated equation

[@mokasaam1819]

Thanks helped for my exam

[@LuciePriscaNgaAteba]

And if the point is not given ??

[@EdwardHon-ev2jk]

❤❤❤❤❤

[@tarirokatengezah7101]

Thnk you so much

[@favouraleph829]

Thanks

[@nomagugumhlanga-g7]

Hoe can l follow you maam

[@IgweDaniel-ft9bq]

Thanks

[@RickGrimessssss]

THANKS