# Find the derivative of the function

#### shamieh

##### Active member
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

Staff member
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
so I let $$\displaystyle u = a^3 + x^3$$

and I got:

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

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#### MarkFL

Staff member
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
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

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#### MarkFL

Staff member
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
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
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

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

Staff member
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!

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
Very nicely done, Shamieh!

Yes, that's correct!

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.