Maths Help (1 Viewer)

[marcquelle]

1) d/dx 2x-3/5-4x
2) vu'-uv'/v^2
3) 2(5-4x)+4(2x-3)/(5-4x)^2
4) 10-8x+8x-12/(5-4x)^2
5) -2/(5-4x)^2

then i get stuck or is that the answer someone help please.

Namu please help!

 

[bubblesss]

hey whats the question? differentiating these?

 

[lyounamu]

1) d/dx 2x-3/5-4x
2) vu'-uv'/v^2
3) 2(5-4x)+4(2x-3)/(5-4x)^2
4) 10-8x+8x-12/(5-4x)^2
5) -2/(5-4x)^2

then i get stuck or is that the answer someone help please.

Namu please help!

d/dx ((2x-3)/(5-4x)) = ((5-4x) . 2 - (2x-3) . -4) / (5-4x)^2
= (10-8x + 8x - 12)/(5-4x)^2
= -2/(5-4x)^2

Yeah, you are right.

 

[kezzaonline]

to my knowledge, that is the answer...

 

[lyounamu]

im not namu but i think i can help
it should be a -4 not +4

It should be +4 because minus times minus is positive.

 

[bubblesss]

yepp ur right.

 

[bubblesss]

It should be +4 because minus times minus is positive.

yeah got a bit muddled with the signs. sry.

 

[marcquelle]

ok thanks guys it just didn't look right

 

[lyounamu]

ok thanks guys it just didn't look right

Yeah some maths questions' answers don't look right. But given the nature of that topic, that kind of answer is really common.

 

[marcquelle]

Yeah some maths questions' answers don't look right. But given the nature of that topic, that kind of answer is really common.

ok thanks *memo to self trust gut instinct with calculus answers* lols

 

[marcquelle]

Find the Equation of the tangent to the parabola x^2=16y at the point (8,4).

This is my solution & answer:

1) y=x^2/16
2) dy/dx = 2x/16 = x/8
3) x=8, dy/dx = 1
Therefor m = 1
4) y-y1=m(x-x1)
5) y-4=1(x-8)
Therfore x-y-4=0

that is my answer should i trust my instincts again or am i completly of track

 

[lyounamu]

Find the Equation of the tangent to the parabola x^2=16y at the point (8,4).

This is my solution & answer:

1) y=x^2/16
2) dy/dx = 2x/16 = x/8
3) x=8, dy/dx = 1
Therefor m = 1
4) y-y1=m(x-x1)
5) y-4=1(x-8)
Therfore x-y-4=0

that is my answer should i trust my instincts again or am i completly of track

100% correct.

 

[tacogym27101990]

Find the Equation of the tangent to the parabola x^2=16y at the point (8,4).

This is my solution & answer:

1) y=x^2/16
2) dy/dx = 2x/16 = x/8
3) x=8, dy/dx = 1
Therefor m = 1
4) y-y1=m(x-x1)
5) y-4=1(x-8)
Therfore x-y-4=0

that is my answer should i trust my instincts again or am i completly of track

nah thats completely correct
need to trust yourself

 

[marcquelle]

aghhhhhhhhh

i hate myself

thank you so much