MHB Find the derivative of the function

[shamieh]

Find the derivative of the function.

$$f(x) = cos(a^3 + x^3)$$

so do I want to distribute and say

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

 

[MarkFL]

The cosine function is not linear, that is to say:

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

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:

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

Can you proceed?

 

[shamieh]

so I let $$ u = a^3 + x^3$$

and I got:

$$y'() = cos(u)$$
$$y'() = -sin(u)$$
$$y'() = -sin(u) * (a^3 + x^3)' .. (f'(g(x) * g'(x))$$
$$y'() = -sin(u) * 3a^2 + 3x^2$$
$$y'() = -sin(a^3 + x^3) * 3a^2 + 3x^2 ?$$

 

[MarkFL]

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:

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

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

 

[shamieh]

well actually I'm taking $$\frac{d}{dy}$$ ( 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 $$y = cos(a^3 + x^3) $$so how do i know what to say like $$y'(whatgoeshere)$$ = ? How do i know whether its x or a. I believe i am taking the $$\frac{dy}{dx}$$ derivative of $$y$$ in respect of $$x$$ if I am saying that right

 

[MarkFL]

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

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

 

[topsquark]

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 believe i am taking the dy/dx derivative of y in respect of x if I am saying that right

Let's see if this helps:
[math]y = cos \left ( a^3 + x^3 \right )[/math]

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

Does that clear up some of the notation for you?

-Dan

 

[shamieh]

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

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

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

 

[MarkFL]

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

You actually want to write:

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

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]

Oh I see now! Since a is constant it goes away, so i have:

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

 

[MarkFL]

Oh I see now! Since a is constant it goes away, so i have:

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

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 $$\sin(x)$$

but

sin(x) gives $$sin(x)$$

 

[DreamWeaver]

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 $$\sin(x)$$

but

sin(x) gives $$sin(x)$$

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 $$\text{mushroom}(x)$$

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

and, less whimsically,

\text{Li}_2(x) shows as $$\text{Li}_2(x)$$
I use this sort of stuff all the time... Especially that mushroom function. (Bandit)