# Is there a Second derivative for 5In(2x) +2x -1

Can someone clarify for me, please?

using the log rules and d/dx In (kx) rule.

we take the five out

5 x d/dx In(2x) + x^2-1

= 5 x d/dx In 2 + In x + x^2 -1

= 5 x 0 + 1/x +2x

= 5/x +2x. I don't know if I have written the notation correctly.

second derivative= -5/x^2 + 2 ? Mathway, informs me that it doesn't exist.
Original post by KingRich
Can someone clarify for me, please?
using the log rules and d/dx In (kx) rule.
we take the five out
5 x d/dx In(2x) + x^2-1
= 5 x d/dx In 2 + In x + x^2 -1
= 5 x 0 + 1/x +2x
= 5/x +2x. I don't know if I have written the notation correctly.
second derivative= -5/x^2 + 2 ? Mathway, informs me that it doesn't exist.
Youve some missing brackets etc in your working and not sure whether its 2x or x^2 originally, but youre basically right. The first derivative of ln(x) is 1/x and its second derivative is -1/x^2
Original post by mqb2766
Youve some missing brackets etc in your working and not sure whether its 2x or x^2 originally, but youre basically right. The first derivative of ln(x) is 1/x and its second derivative is -1/x^2

Apologise. I’m not sure how to use the fancy powers on desktop outside of Microsoft word so it’s a little confusing…

Original function is: 5In(2x)+x²-1 ,

So the brackets missing are for 5 x d/dx (In (2) + In (x)) + x²-1…..

I suppose 5(0+1/x) + 2x, too

In that case when inputting into online calculators, why do you think it might be showing incorrect. For example I tried math papa and it came back with 10in + 2x as the first derivative. Perhaps I’m inputting it incorrectly?
Original post by KingRich
Apologise. I’m not sure how to use the fancy powers on desktop outside of Microsoft word so it’s a little confusing…
Original function is: 5In(2x)+x²-1 ,
So the brackets missing are for 5 x d/dx (In (2) + In (x)) + x²-1…..
I suppose 5(0+1/x) + 2x, too
In that case when inputting into online calculators, why do you think it might be showing incorrect. For example I tried math papa and it came back with 10in + 2x as the first derivative. Perhaps I’m inputting it incorrectly?

https://www.wolframalpha.com/input?i=second+derivative+5ln%282x%29%2Bx%5E2-1

You can either expand ln(2x) and differentiate ln(2)+ln(x) or use the usual rule
d ln(f(x))/dx = f'(x)/f(x)
so here its 2/(2x) = 1/x
Original post by mqb2766
https://www.wolframalpha.com/input?i=second+derivative+5ln%282x%29%2Bx%5E2-1
You can either expand ln(2x) and differentiate ln(2)+ln(x) or use the usual rule
d ln(f(x))/dx = f'(x)/f(x)
so here its 2/(2x) = 1/x

So weird. When using the desktop, mine doesn’t acknowledge the function.

I realised the issue.

It doesn’t recognise the type command, so have to use the functions given on their system
(edited 1 month ago)