You are not logged in.
would it be the same way differentaiting the function above as this (3x + 5)^10 function?
f(x) = x^4-4x^2+4 g(x) = x^6-4x^3+4x^2
d [f(x)g(x)] = (x^4-4x^2+4)(6x^5-12x^2+8x) + (4x^3-8x)(x^6-4x^3+4x^2)
i think i know what to do now
x^4-4x^2+4 ? am i getting close?
ooo
how about this?
(x^2-2)(x^2-2)
2x^4-2x^2-2x^2+4
2x^4-4x^2+4
f(x)= (x^2-2)^2 f(g)= (x^3-2x)^2
= (2x^4-4) = (2x^5-8x^2)
(2x^4-4)(10x^4-16x) + (2x^5-8x^2)(8x^3)
hey,
Do i expand the brackets before i differentiate?
1) (x² + 2)² ÷ (x² + x-¹)³
2) (x² + 1)² (x³ - 2x)²
some how i got y=0
how do i find y and b ? plug -1 into the derivative ?
like this?
3x+3=0 x-3=0
x=-1 x=3
ops (3x+3)(x-3)
(3x^2+3)(x-3)
3x^2-6x=9
Something like this?
9 = (x-2)^2 + 2(x+1)(x-2)
9 = x^2-2x-2x+4+2x^2-4x+2x-4
3x^2=6x=9
do i use y=4 to find the x co-ordinate?
Am i finding the x co-ordinate of a ponit where the tangent is parallel to another tangent?
ah gotcha
oooo found my mistake 6x3=18 not 24 ops
4 = (3-2)^2 + 2(3+1)(3-2)
4 = (3-2)(3-2) + (6+2)(3-2)
9-6-6+4+24-12+6-4-4= 11
where did that equation come from? wasnt this the first derivative f'(X) = (x + 1)2(x+1) + (x-2)^2
umm where doe the 4 go?
plug (3,4) into f'(x) = (x +1)2(x+1) + (x-2)^2?
sry just thinking about it, i dont really get the question either the "tangent is parallel to the tangent at the point (3,4)" confuses me dont worry about i will get my lecturer the explain thx for the help.
umm the whole question goes like this,
A) find the x co-ordinate of the points on the curve Y = (x+1)(x-2)^2 at which the gradient is zero. test whether the points are minima or maxima. this part with your help is done. thx agian
B) Also find the x co-ordinate of the point at which tangent is parallel to the tangent at the point (3,4)