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Find the derivative of the function

shamieh

Active member
Sep 13, 2013
539
Find the derivative of the function.

\(\displaystyle f(x) = cos(a^3 + x^3)\)

so do I want to distribute and say

\(\displaystyle f'(x) = cosa^3 + cosx^3\) and then use chain rule or...How exactly should I go about solving this particular problem?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The cosine function is not linear, that is to say:

\(\displaystyle \cos(\alpha\pm\beta)\ne\cos(\alpha)\pm\cos(\beta)\)

You could use the angle-sum identity for cosine, but this is not necessary.

What you want to do is apply the differentiation rule for cosine and the chain rule:

\(\displaystyle \frac{d}{dx}\left(\cos\left(a^3+x^3 \right) \right)=-\sin\left(a^3+x^3 \right)\frac{d}{dx}\left(a^3+x^3 \right)\)

Can you proceed?
 

shamieh

Active member
Sep 13, 2013
539
so I let \(\displaystyle u = a^3 + x^3\)

and I got:

\(\displaystyle y'() = cos(u)\)
\(\displaystyle y'() = -sin(u)\)
\(\displaystyle y'() = -sin(u) * (a^3 + x^3)' .. (f'(g(x) * g'(x))\)
\(\displaystyle y'() = -sin(u) * 3a^2 + 3x^2\)
\(\displaystyle y'() = -sin(a^3 + x^3) * 3a^2 + 3x^2 ?\)
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Are you told whether $a$ is a constant, or if it is a function of $x$? I would assume it is a constant unless told it depends on $x$.

If $a$ is a function of $x$ as you are treating it, then there are a few issues I wish to address:

i) To complete the use of the chain rule, you want to append the derivative of $a$ with respect to $x$ to the term involving $a$.

ii) You should enclose the derivative of the argument of the trig. function within parentheses.

Hence, you would have:

\(\displaystyle \frac{d}{dx}\left(\cos\left(a^3+x^3 \right) \right)=-\sin\left(a^3+x^3 \right)\frac{d}{dx}\left(a^3+x^3 \right)=-\sin\left(a^3+x^3 \right)\left(3a^2\frac{da}{dx}+3x^2 \right)=\)

\(\displaystyle -3\sin\left(a^3+x^3 \right)\left(a^2\frac{da}{dx}+x^2 \right)\)
 

shamieh

Active member
Sep 13, 2013
539
well actually I'm taking \(\displaystyle \frac{d}{dy}\) ( when i say that I am guessing, i am not really sure) I i meant to say y'. I'll fix that. All it said was differentiate \(\displaystyle y = cos(a^3 + x^3) \)so how do i know what to say like \(\displaystyle y'(whatgoeshere)\) = ? How do i know whether its x or a. I beleive i am taking the \(\displaystyle \frac{dy}{dx}\) derivative of \(\displaystyle y\) in respect of \(\displaystyle x\) if im saying that right
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Okay, I would assume then that $a$ is a constant...otherwise it should be written as $a(x)$. That being the case, what is:

\(\displaystyle \frac{d}{dx}\left(a^3+x^3 \right)\) ?
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,140
well actually I'm taking d/dy ( when i say that I am guessing, i am not really sure) I i meant to say y'. I'll fix that. All it said was differentiate y = cos(a^3 + x^3) so how do i know what to say like y'(whatgoeshere)' = ? How do i know whether its x or a. I beleive i am taking the dy/dx derivative of y in respect of x if im saying that right
Let's see if this helps:
\(\displaystyle y = cos \left ( a^3 + x^3 \right )\)

and you need to find:
\(\displaystyle \frac{dy}{dx} = \frac{d}{dx} \left [ cos \left ( a^3 + x^3 \right ) \right ]\)

Does that clear up some of the notation for you?

-Dan
 

shamieh

Active member
Sep 13, 2013
539
Okay, I would assume then that $a$ is a constant...otherwise it should be written as $a(x)$. That being the case, what is:

\(\displaystyle \frac{d}{dx}\left(a^3+x^3 \right)\) ?
\(\displaystyle \frac{d}{dx}(a^3 + x^3) = \frac{d}{dx}(3x^2)\) ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
\(\displaystyle \frac{d}{dx}(a^3 + x^3) = \frac{d}{dx}(3x^2)\) ?
You actually want to write:

\(\displaystyle \frac{d}{dx}\left(a^3+x^3 \right)=3x^2\)

The right side is the result of taking the derivative with respect to $x$. Now, do you know what to do with this result?
 

shamieh

Active member
Sep 13, 2013
539
Oh I see now! Since a is constant it goes away, so i have:

\(\displaystyle -sin(u) * 3x^2\) which becomes \(\displaystyle -3x^2 sin(u)\) which becomes \(\displaystyle y'(x)\) or \(\displaystyle \frac{dy}{dx}\) = \(\displaystyle -3x^2 sin(a^3 + x^3)\)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Oh I see now! Since a is constant it goes away, so i have:

\(\displaystyle -sin(u) * 3x^2\) which becomes \(\displaystyle -3x^2 sin(u)\) which becomes \(\displaystyle y'(x)\) or \(\displaystyle \frac{dy}{dx}\) = \(\displaystyle -3x^2 sin(a^3 + x^3)\)
Yes, that's correct! (Sun)

I want to offer a $\LaTeX$ tip...

Precede predefined functions such as the trigonometric and logarithmic functions with a backslash so that their names are not italicized. This way they are distinguished from a string of variables. For example:

\sin(x) gives \(\displaystyle \sin(x)\)

but

sin(x) gives \(\displaystyle sin(x)\)
 

DreamWeaver

Well-known member
Sep 16, 2013
337
Very nicely done, Shamieh! :D


Yes, that's correct! (Sun)

I want to offer a $\LaTeX$ tip...

Precede predefined functions such as the trigonometric and logarithmic functions with a backslash so that their names are not italicized. This way they are distinguished from a string of variables. For example:

\sin(x) gives \(\displaystyle \sin(x)\)

but

sin(x) gives \(\displaystyle sin(x)\)

Following on from that excellent Latex advice from Mark, if you want to introduce a named function that doesn't have it's own bespoke code, you can still easily get the same effect by using the following tags

\text{}


For example...

\text{mushroom}(x) shows as \(\displaystyle \text{mushroom}(x)\)

\text{Pie}(\pi) shows as \(\displaystyle \text{Pie}(\pi)\)

and, less whimsically,

\text{Li}_2(x) shows as \(\displaystyle \text{Li}_2(x)\)



I use this sort of stuff all the time... Especially that mushroom function. (Bandit)