How to get the second derivative of this function ?

[AbuYusufEg]

Let a function v in one variable, say u
u is a function also, but in two variables, say x & y

for the first derivative of v, i did the following:
\frac{\partial v}{\partial x} = \frac{d v}{d u} . \frac{\partial u}{\partial x}

And the resutlt is:
\frac{\partial v}{\partial x} = cos(u) . u_x

Note: u_x is so hard to get, may be impossible without computer.

But, if i want the second derivative of v, How to ?

\frac{\partial^2 v}{\partial x^2} = ?

i think that the way for getting it is Differentiating the result, means differentiating: cos(u) . u_x
But how to also ?

 

[tiny-tim]

Welcome to PF!

Hi AbuYusufEg! Welcome to PF!

Just treat cos(u) as your new v, and use the product rule.

(there'll be a (u_x)² in it … was that putting you off? )

Have a go!

 

[AbuYusufEg]

i'll give a try !, tell me if any thing wrong ..
\frac{\partial^2 v}{\partial x^2} = -sin(u).u_x + cos(u).u_{xx}
is that correct ? and where is the (u_x)^2 you talk about ? and why should it be here ?

 

[tiny-tim]

The second term is correct, but the first one should be -sin(u).u_x² …

can you see why?

 

[AbuYusufEg]

Sorry, But No !

 

[tiny-tim]

ok … you have to use the product rule on cos(u).u_x …

so it's cos(u).(∂/∂x(u_x)) + (∂/∂x(cos(u)).u_x

= cos(u).u_xx + … ?

 

[HallsofIvy]

v_x= cos(u)u_x
use the product rule:

v_{xx}= (cos(u))_x u_x+ cos(u) u_{xx}
And (cos(u))_x= -sin(u) u_x by the chain rule. Because of the "u_x" in that,
(cos(u))_x u_x= (-sin(u) u_x)u_x= -sin(u) u_x^2.

 

[AbuYusufEg]

ok … you have to use the product rule on cos(u).u_x …

so it's cos(u).(∂/∂x(u_x)) + (∂/∂x(cos(u)).u_x

= cos(u).u_xx + … ?

v_x= cos(u)u_x
use the product rule:

v_{xx}= (cos(u))_x u_x+ cos(u) u_{xx}
And (cos(u))_x= -sin(u) u_x by the chain rule. Because of the "u_x" in that,
(cos(u))_x u_x= (-sin(u) u_x)u_x= -sin(u) u_x^2.

it's clear now !, good.