SOLUTION: Hi, i'm not sure which topic it should go in but anyway... i am struggling to find out how to get the answer to this. 'Find the gradient of the following curves at the given

Algebra ->  Graphs -> SOLUTION: Hi, i'm not sure which topic it should go in but anyway... i am struggling to find out how to get the answer to this. 'Find the gradient of the following curves at the given       Log On


   

Question 159386This question is from textbook pure core maths1 & 2
: Hi, i'm not sure which topic it should go in but anyway...
i am struggling to find out how to get the answer to this.
'Find the gradient of the following curves at the given point'
y = 3*rootx(x-2)^2
at the point where x = 3
i have got the equation to
y = 3*rootx(x^2 -4x + 4)
i then have to find the gradient (dy/dx) and substitute in x (3)
but i can't get the right answer, which the book says is 29*root3
thanks in advance. This question is from textbook pure core maths1 & 2

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
First off, let's do a substitution.
Let
u=x%28x-2%29%5E2
then
y+=+3%2Au%5E%281%2F2%29
dy%2Fdx=3%2A%281%2F2%29u%5E%28-1%2F2%29%2A%28du%2Fdx%29
Now we can go back and substitute,
Let's look at u,
u=x%28x-2%29%5E2
u=x%28x%5E2-4x%2B4%29
u=x%5E3-4x%5E2%2B4x
du%2Fdx=3x%5E2-8x%2B4
u%283%29=3%283-2%29%5E2=3%281%29=3
%28du%2Fdx%29%283%29=3%283%29%5E2-8%283%29%2B4=27-24%2B4=7
Substitute all those values into dy/dx,
%28dy%2Fdx%29=3%2A%281%2F2%29u%5E%28-1%2F2%29%2A%28du%2Fdx%29
%28dy%2Fdx%29%283%29=3%2A%281%2F2%29%283%29%5E%28-1%2F2%29%2A%287%29
%28dy%2Fdx%29%283%29=%2821%2F2%29%283%29%5E%28-1%2F2%29=%2821%2F2%29%281%2Fsqrt%283%29%29
%28dy%2Fdx%29%283%29=%2821%2F2%29%28sqrt%283%29%2F3%29=%287sqrt%283%29%29%2F2
Is this the answer you got??
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We can check the answer by calculating dy/dx near 3 using x and y values.
Let's use
x=2.99, then y=+3%2Asqrt%282.99%282.99-2%29%5E2%29=+5.13561
x=3.00, then y=+3%2Asqrt%283%283-2%29%5E2%29=+5.196152
x=3.01, then y=+3%2Asqrt%283.01%283.01-2%29%5E2%29=+5.256854
then we can calculate the derivative on the left of 3 and the right of 3 and take the average.
(dy/dx)left=y(3.00)-y(2.99))/(3.00-2.99)=(5.196152-5.13561)/(0.01)=6.054233
(dy/dx)right=y(3.01)-y(3.00))/(3.01-3.00)=(5.256854-5.196152)/(0.01)=6.07011
We take the average of those two and get
(dy/dx)(3)=6.062
Let's look at the value we calculated,
%287%2Asqrt%283%29%29%2F2=12.12436%2F2=6.062178
Looks like the two match.
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Meanwhile, the answer you provided doesn't look close.
29%2Asqrt%283%29=+50.22947 is not near the answer we calculated and verified.
Check the answer again.