Easy differentiation maths question my mind is currently blank on (1 Viewer)

[StaceyK293920]

Differentiate from first principles to find the gradient and tangent to a curve y=x^2+x at the point (2,6)

I get confused when I try to modify f(x+h) which annoys me. Any easy way to do it?

Ty

 

[onebytwo]

Differentiate from first principles to find the gradient and tangent to a curve y=x^2+x at the point (2,6)

I get confused when I try to modify f(x+h) which annoys me. Any easy way to do it?

Ty

replace every x you see in the given equation with (x+h).

ie the equation becomes f(x+h) = (x+h)^2 + (x+h)

try to continue....

 

[Aerath]

f(x) = x^2 + x
f(x+h) = (x+h)^2 + (x+h)
= x^2 + 2xh + h^2 + x + h

Differentiation from first principles:
f'(x) = limit of h towards 0 of [f(x+h)-f(x)] / h
= limit of h towards 0 of [(x^2 + 2xh +h^2 + x + h) - (x^2 +x)] / h
= limit of h towards 0 of [2xh + h^2 +h] / h
= limit of h towards 0 of [2x + h + 1]
= 2x + 1

Therefore y' = 2x + 1
y' (x = 2) = 2(2) +1 = 5
Therefore gradient at x = 2, is m = 5.


Point gradient formula:
y - y1 = m(x - x1)
y - 6 = 5(x-2)
y -6 = 5x - 10
y = 5x - 4

Hopefully I made no mistakes. But the quicker way to differentiate (which you can't do in this question, because it specifically asks for "from first principles"), (x^n)' = nx^(n-1)

 

[tacogym27101990]

u have

lim [(x+h)^2 +(x+h) - (x^2 +x)]/h
h>0

lim [x^2 + 2xh +h^2 +x+h - x^2 - x]/h
h>0

lim [h(2x +1 +h)]/h
h>0

lim 2x + 1 +h
h>0

=2x +1

 

[onebytwo]

good work aerath...but notice it should be h-->0 not infinity.

 

[tacogym27101990]

f(x) = x^2 + x
f(x+h) = (x+h)^2 + (x+h)
= x^2 + 2xh + h^2 + x + h

Differentiation from first principles:
f'(x) = limit of h towards infinity of [f(x+h)-f(x)] / h
= limit of h towards infinity of [(x^2 + 2xh +h^2 + x + h) - (x^2 +x)] / h
= limit of h towards infinity of [2xh + h^2 +h] / h
= limit of h towards infinity of [2x + h + 1]
= 2x + 1

Therefore y' = 2x + 1
y' (x = 2) = 2(2) +1 = 5
Therefore gradient at x = 2, is m = 5.


Point gradient formula:
y - y1 = m(x - x1)
y - 6 = 5(x-2)
y -6 = 5x - 10
y = 5x - 4

Hopefully I made no mistakes. But the quicker way to differentiate (which you can't do in this question, because it specifically asks for "from first principles"), (x^n)' = nx^(n-1)

its the limit as h approaches 0
if it was infinity you wouldn't have a tangent

 

[Aerath]

Oh yeah, what you said.
Thanks for the find, and I've edited my post above (so I won't look like an idiot).

 

[Aplus]

Oh yeah, what you said.
Thanks for the find, and I've edited my post above (so I won't look like an idiot).

Might want to delete this post as well

 

[StaceyK293920]

Thanks all <3.

 

[Continuum]

Might want to delete this post as well

Don't quote him, he won't be able to delete your post.