Question

Find the coordinates of the stationary point of the curve y=ln(x²-6x+10).

Answer 1

all stat pts-> dy/dx = 0

Answer 2

i.e. differentiate your function and solve

Answer 3

To differentiate log e,
Take whatever is in the bracket,
put it on the bottom of a fraction (denominator),
differentiate the bottom of the fraction
then place the differentiated expression on the top of the fraction (numerator)

Answer 4

\[\frac{ dy }{ dx } \ln(x ^{2} - 6x +10) = \frac{ 1 }{ x ^{2} - 6x + 10 }\]

Answer 5

@Skaematik ?

Answer 6

@curiousshubham Almost.
You need to use the chain rule to differentiate this.\[{d \over dx}(f(g(x)))=f^{\prime}(g(x))g^{\prime}\]\[{d \over dx}\ln(x^2-6x+10)\] Let \[f(x)=ln(x)\] and \[g(x)=x^2-6x+10\]
Find \[f^{\prime}={d \over dx}ln(x)\]\[g^{\prime}={d \over dx}(x^2-6x+10)\]
And substitute into the equation up the top.

Answer 7

@PhoenixFire I got
\[\frac{ 1 }{ x ^{2} - 6x +10 }\]

Answer 8

that's df/dx. you need to multiply by the derivative of the inside function (dg/dx)

Answer 9

y = ln(f(x))
y' = f'(x)/f(x) , right?

Answer 10

yup.

Answer 11

So you get
\[{d \over dx}\ln(x^2-6x+10)={{2x-6} \over {x^2-6x+10}}\]

Answer 12

yup!

Answer 13

@PhoenixFire Equatting it with 0?

Answer 14

Yeah, equate to 0 and solve for x. for each value of x plug it into the original equation to get your y. then you will have your coordinates.