That’s actually a definition. Basically, differentiating y once is dy/dx, and differentiating it once more gives us d2y/dx2.

The statement simply means “differentiating dy/dx once more”.

So you don’t actually differentiate dy/dx

As in, if I differentiate dy/dx with respect to x,

d/dx(dy/dx)

= d/dx (dy(dx)^-1)

= dy . d/dx((dx)^-1)

= dy . -(dx)^-2(d)

= d^2y(-dx)^-2

= - d^2y/d^2x^2

1/5x is (5x)^-1 or 5x^-1 ?

No, “dy/dx” is just a terminology and it does not exactly mean “dy divided by dx”, so we cannot exactly use “quotient rule” on it.

1/(5x), with the entire 5x being in the denominator, is (5x)^-1.

But for differentiation purposes, you should rewrite it as 1/5 times 1/x, or 1/5 times x^-1.

Oh okok

But if it is (5x)^-1, would the chain rule apply?

Yes of course, because the bracketed term is not the basic term “x”, but it’s “5x”.

d/dx ( (5x)^-1)

= -(5x)^-2(5)

= -5(5x)^-2

= -5/(5x)^2

= -1/5x^2

So as long as there is more than x in the bracket, chain rule apply

Thank you!!!

Typically, yes, more than 1 bracket involves chain rule

Why is it when

d^2y/dx^2 < 0 is maximum point

d^2y/dx^2 > 0 is minimum point

Whenever we have a maximum point P, a curve will have a positive gradient before P, a zero gradient at P and a negative gradient at P. Throughout the motion, the gradient keeps decreasing.

A graph which always decreases will have a negative gradient. Similarly, if the gradient of a curve (dy/dx) is always decreasing, ITS gradient (d2y/dx2) will be negative.

As such, d2y/dx2 < 0 for a maximum point.

A similar idea applies for the minimum point.

Thanks!!

Thing to take note :


① If given that d²y/dx² < 0 when dy/dx = 0 , it implies a maximum point.

But this does not mean that a maximum point always has d²y/dx² < 0

Example : y = -x⁴ , at x = 0

dy/dx = -4x³
d²y/dx² = d/dx (dy/dx)
= -12x²

When x = 0,

d²y/dx² = -12(0²) = 0

But if we sketch the graph of y = -x⁴, it has the same shape as y = -x², and the maximum point is at x = 0.

Likewise,

If given that d²y/dx² > 0 when dy/dx = 0 , it implies a minimum point.

But this does not mean that a minimum point always has d²y/dx² > 0



② A stationary point of inflection always has d²y/dx² = 0 and dy/dx = 0

The gradient goes from one sign before this point, then to 0 at this point, then back to the same sign after this point.

e.g positive to 0, then to positive again.
Or
negative to 0, then back to negative again.


BUT, the gradient of the gradient however, changes from 1 sign to 0 , then to another sign.

eg. negative to 0, then to positive. Jt keeps increasing as x increases.
Or
Positive to 0, then to negative. It keeps decreasing as x increases.




But, if given or found that d²y/dx² = 0, it does not necessarily imply a stationary point (UNLESS, dy/dx is also found to be = 0)


If given/found that d²y/dx² = 0 when dy/dx = 0, this also does not necessarily mean that it's an stationary point of inflection.

(Refer to the very first example y = -x⁴ above)


We will need to go back to the first derivative test when we get d²y/dx² = 0 for the second derivative test.

In short ,

If given/found that :


dy/dx = 0 → implies stationary point, vice versa

Inflection point → implies d²y/dx² = 0
But d²y/dx² = 0 does not necessarily imply inflection point.


dy/dx = 0 and d²y/dx² < 0 → maximum point

dy/dx = 0 and d²y/dx² > 0 → minimum


dy/dx = 0 and d²y/dx² = 0 → use first derivative test to check.

(it could be either maximum, minimum or stationary point of inflection))

Wow thanks for the detailed explanation!!
Does this mean that first derivative test is more accurate then the second derivative test?
So when do I use second derivative test over the first one?

Usually second derivative test is preferred when the function dy/dx can be differentiated easily.

First derivative test is still the standard way since we are “inspecting” the curve on a point by point basis.

Got it thanks!

https://internal.ncl.ac.uk/ask/numeracy-maths-statistics/core-mathematics/calculus/stationary-points.html

This may be helpful.

Using the second derivative test usually involves some sort of foresight. i.e if you can see that the dy/dx is easily differentiated and will lead to negative or positive d²y/dx²

Eg. x⁴ + 4x + 1

dy/dx = 4x³ + 4


Before writing anything else, I already can forsee that d²y/dx² has only terms in x². Since dy/dx = 0 at x= -1,

x ≠ 0 means x² is positive and therefore d²y/dx² won't be 0. I can proceed to do 2nd derivative test.

I see. Thanks!

QN, as long as my expressions are not of the form

(x + a)^n

where n > 2

second derivative test should work well (not sure if there are other exceptions as well).

For example, if I have

(x + 1)^8 + (x + 1)^7

then second derivative test would fail, because the “x + 1” would survive later on (we needed to set x + 1 to 0 for the turning point).

But if our expression is say

y = 3x^2 - ln x

then second derivative test works well.

Ok!

For y = (x + 1)^8 + (x + 1)^7,

There are two stationary points. It depends on which is to be checked for maximum/ minimum/point of inflection.

Second derivative works for when x = -15/8, doesn't work for x = -1

Exercise discretion where necessary

So do you mean
When y= (x+1)^8 + (x+1)^7
d/dx = 8(x+1)^7 + 7(x+1)^6

Using second derivative test,
d^2/dx^2
= d/dx(dy/dx)
= 56(x+1)^6 + 42(x+1)^5

So second derivative will not work because the is two minimum/maximum/stationary point of inflexion?

if you sub x = -1 into d²y/dx², you will get 0.

So you can't tell the nature of the stationary point at x = -1. You'll need to do first derivative test.

But if the question had asked you to check for x = -15/8 (the other stationary point),

d²y/dx² = 56(-15/8+1)^6 + 42(-15/8 + 1)^5

= 117649/32768
= 3 19345/32768

≈ 3.59 > 0 (minimum)

Alternatively,


56(-15/8+1)^6 + 42(-15/8 + 1)^5

= 56(-⅞)^6 + 42(-⅞)^5
= (-⅞)^5 (56(-⅞) + 42)
= (-⅞)^5 (-49 + 42)
=(-⅞)^5 (-7)

Since I know both terms are negative, the product = d²y/dx² must be positive without even evaluating it.

How did you get -1?
I understand the -15/8 part when the d/dx=0

There are two stationary points. One is at x = -1. The other is at x = -15/8

You can see the following working :

When dy/dx = 0,

8(x + 1)^7 + 7(x + 1)^6 = 0

Factorise,

(x + 1)^6 ( 8(x + 1) + 7) = 0

(x + 1)^6 (8x + 15) = 0


(x + 1)^6 = 0 or 8x + 15 = 0

x + 1 = 0 or 8x = -15

x = -1 or x = -15/8


Realise that (x + 1)^6 = 0 is just repeated roots.

So if the question had asked to check the nature of the point at x = -1,


d²y/dx²

= 56(-1 + 1)^6 + 42(-1 + 1)^5

= 56(0)^6 + 42(0)^5

= 56 x 0 + 42 x 0

= 0

Ohh okok, thanks:)