# Second derivative

## Homework Statement

Find the second derivative of 3sec$$\sqrt{x}$$

## Homework Equations

derivative of sec u = secu(tanu)*u'

product rule- (F*G)'=f'*g+g'*f
quotient rule- (F/G)'=(f'*g-f*g')/g^2

## The Attempt at a Solution

So i did the first derivative of it, getting 3sin$$\sqrt{x}$$/{2$$\sqrt{x}$$*cos^2$$\sqrt{x}$$}

After this point, I started to simply take this derivative again. However, my paper says to find d^2y/d^2x, which is confusing to me since there is no y term in this, so is it just simply taking the derivative twice?

Related Calculus and Beyond Homework Help News on Phys.org
Dick
Homework Helper
Yes, it's simply taking the derivative twice. It's d/dx(dy/dx).

I would say yes .. take the derivative again .. I think d^2y/d^2x, should read d2y/dx2 ...

I have the first derivative to be f' = (3/2) ((sec$$\sqrt{}x$$)$$\cdot$$(tan$$\sqrt{}x$$))/$$\sqrt{}x$$

Danger,
If you were to simplify that to just sin and cos, then it would be the same as mine. I did this so the derivatives of the trig functions in the quotient rule portion would be easier (just -sin or cos).

try this one
substitute for x^1/2

let y = 3sec(x^1/2)
let u = x^1/2

thus
y = 3sec(u)
y' = 3sec(u)tan(u)
now second derivative is
y'' = 3[sec(u)tan(u).tan(u) + sec^2(u)sec(u)]
y'' = 3[sec(u)tan^2(u) + sec^3(u)]

substuting back for u

y'' = 3[sec(x^1/2)tan^2(x^1/2) + sec^3(x^1/2)]

Last edited:
whoaaaaaaa,
Electrophysics, I am not that far into derivitives to know whether or not that is an application, but are you saying I can do that? or are you like saying, maybe you could do that. That seems so much easier than what i did (which took about a page and a half of paper).

try this one
substitute for x^1/2

let y = 3sec(x^1/2)
let u = x^1/2

thus
y = 3sec(u)
y' = 3sec(u)tan(u)
now second derivative is
y'' = 3[sec(u)tan(u).tan(u) + sec^2(u)sec(u)]
y'' = 3[sec(u)tan^2(u) + sec^3(u)]

substuting back for u

y'' = 3[sec(x^1/2)tan^2(x^1/2) + sec^3(x^1/2)]

And, wouldn't the first derivative of 3secu be
3sec(u)tan(u)(u')??? often it doens't matter because it is just an x or whatever, but since squareroot of x would have a derivative, doens't that come into play.?

Electro physics - u have have calculated d^2y/du^2
ie: by the second derivative of x^1/2.

This is a simple chain rule application.

hesbon,
If i take the answer that electro got and multiply it by f'' of x^1/2 i'll get the correct answer?

Also, if i were to take the first derivative, and simpify it so that it is in sin and cos, i will still get the same derivative as if i keep it in secants and tangents correct?

U-substitution just makes it more complicated.

Just take the first derivative.

that would be (1/2)*3 * sec(sqrt[x]) * tan(sqrt[x])

So you have sec * tan * 3/2

Take the derivative again using the chain rule.

Mark44
Mentor
And, wouldn't the first derivative of 3secu be
3sec(u)tan(u)(u')??? often it doens't matter because it is just an x or whatever, but since squareroot of x would have a derivative, doens't that come into play.?
Yes and yes.
The work posted by ElectroPhysics is wrong if by y' he means dy/dx, and by y'' he means d^2y/dx^2.

If both of these derivatives are with respect to u, then the work is right, but incomplete, as the original problem asked for the first and second derivatives w.r.t. x.
let u = x^1/2

thus
y = 3sec(u)
y' = 3sec(u)tan(u)
now second derivative is
y'' = 3[sec(u)tan(u).tan(u) + sec^2(u)sec(u)]
y'' = 3[sec(u)tan^2(u) + sec^3(u)]

substuting back for u

y'' = 3[sec(x^1/2)tan^2(x^1/2) + sec^3(x^1/2)]
The prime notation is fine as long as it's understood with respect to which variable differentiation occurs. As soon as you introduce a third variable, there's the potential for losing clarity.

Here's the first derivative again, this time in a more explicit form:
dy/du = 3sec(u)tan(u)
dy/dx = dy/du * du/dx = 3sec(sqrt(x))tan(sqrt(x)) * d(sqrt(x))/dx

assuming that he has calculated the derivative with respect to u correctly than ..yes all u have to do is multiple by second derivative of x^1/2.

you can present the initial function in many equivalent trignometric forms
The derivatives of all of them will also be equivalent.
however they may appear to be different, but with a little trig manipulation you will find that they are infact identical.

thank you heshbon!

I dont know how long it would take you/if it is alot to ask, but i was wondering if you could post what you would get if you were to take the derivative of

3sin$$\sqrt{x}$$/(2$$\sqrt{x}$$cos^2$$\sqrt{x}$$)

This is the first derivative of 3sec$$\sqrt{x}$$ after simplifying to sin and cos.

-3(sin$$\sqrt{x}$$/(4x^3/2cos^2$$\sqrt{x}$$)+(sin^2$$\sqrt{x}$$cos$$\sqrt{x}$$)/(2x(cos^2$$\sqrt{x}$$)^2)+cos$$\sqrt{x}$$/(4xcos^2$$\sqrt{x}$$)

wow, that is just about impossible to read, but does that look like it would work?