# Differentiation Rules Question

#### Hckyplayer8

##### New member
Hello. I'm in the portion of the course where you start moving from difference quotients to the various rules.

The specific question I'm working on is

Suppose f(x) = x^3 cos x

(a) Use the differentiation rules to compute f prime (x).
(b) Find an equation for the line tangent to the graph of f at x = π.

Okay so the first part seems pretty simple. Got the power rule down and while the proof of the trig rule was a tad daunting, I'm fine with memorizing that the derivative of cosine is negative sine.

So for part (a) I have 3x^2 - sin x . Looking at that I feel like I am done. The product rules doesn't look like an option cause it's just a single function.

So pending I'm done with (a), (b) looks like it is asking for point slope form. It is here that my understanding breaks down. I'm not even sure where to begin.

#### MarkFL

##### Super Moderator
Staff member
Hello, and welcome to FMH!

You do need to use the product rule here, because you have the product of $$x^3$$ and $$\cos(x)$$.

#### pka

##### Elite Member
Suppose f(x) = x^3 cos x
(b) Find an equation for the line tangent to the graph of f at x = π.
$$\displaystyle f'(\pi)=3(\pi^2)\cos(\pi)-(\pi)^3\sin(\pi)$$
The tangent at $$\displaystyle (\pi,(\pi)^3\cos(\pi))$$ is $$\displaystyle (y-(\pi)^3\cos(\pi))=f'(\pi)(x-\pi)$$

#### Hckyplayer8

##### New member
Hello, and welcome to FMH!

You do need to use the product rule here, because you have the product of $$x^3$$ and $$\cos(x)$$.
Thank you.

The question only gives me f(x). Wouldn't I need g(x) to use the product rule? Or am I reading the definition wrong?

#### Hckyplayer8

##### New member
$$\displaystyle f'(\pi)=3(\pi^2)\cos(\pi)-(\pi)^3\sin(\pi)$$
The tangent at $$\displaystyle (\pi,(\pi)^3\cos(\pi))$$ is $$\displaystyle (y-(\pi)^3\cos(\pi))=f'(\pi)(x-\pi)$$
Thank you. Could you please explain that? Its pretty much point slope form right? y-y1 = m (x-x1)

#### MarkFL

##### Super Moderator
Staff member
Thank you.

The question only gives me f(x). Wouldn't I need g(x) to use the product rule? Or am I reading the definition wrong?
You can thing of it this way:

$$\displaystyle f(x)=u(x)v(x)$$

where:

$$\displaystyle u(x)=x^3$$

$$\displaystyle v(x)=\cos(x)$$

#### pka

##### Elite Member
Could you please explain that? Its pretty much point slope form right? y-y1 = m (x-x1)
Suppose $$\displaystyle f(x) = x^3 \cos( x)$$ Do you know to differentiate a product?
$$\displaystyle f'(x)=3x^2\cos(x)-x^3\sin(x)$$

#### Hckyplayer8

##### New member
You can thing of it this way:

$$\displaystyle f(x)=u(x)v(x)$$

where:

$$\displaystyle u(x)=x^3$$

$$\displaystyle v(x)=\cos(x)$$
Got it. Most of the problems spelled out the second function for me up to this point.

So for (fg)prime of x equaling f prime(x) times g(x) plus f(x) times g prime (x), the problem is really saying differentiate the two "terms" (is that the right verbiage) within f(x) and then multiply the derivatives of terms by the opposing term?

#### Hckyplayer8

##### New member
Suppose $$\displaystyle f(x) = x^3 \cos( x)$$ Do you know to differentiate a product?
$$\displaystyle f'(x)=3x^2\cos(x)-x^3\sin(x)$$
I believe I am following. To differentiate a product you use the product rule which is (fg)prime of x equaling f prime(x) times g(x) plus f(x) times g prime (x).

Derivative of x^3 gets you 3x^2 (f prime) . Multiply that by cos x[g(x]. Then add it to x^3 [f(x)] and the derivative of cos x which is - sin x.

The only thing I don't understand (of what you posted regarding the differentiation) is how bringing the subtraction sign from the sin x to include the x^3 doesn't change what the problem is saying.

For equation of a tangent line part, I don't know how one chooses which portion of the differentiation equation is inserted as the y1,x1 and slope.

#### MarkFL

##### Super Moderator
Staff member
I would say you are being asked to differentiate a function, that is composed of 2 factors. That is, it is the product of 2 factors, and so the product rule is the appropriate tool to apply.

#### Dr.Peterson

##### Elite Member
Derivative of x^3 gets you 3x^2 (f prime) . Multiply that by cos x[g(x]. Then add it to x^3 [f(x)] and the derivative of cos x which is - sin x.

The only thing I don't understand (of what you posted regarding the differentiation) is how bringing the subtraction sign from the sin x to include the x^3 doesn't change what the problem is saying.

For equation of a tangent line part, I don't know how one chooses which portion of the differentiation equation is inserted as the y1,x1 and slope.
That isn't a subtraction sign; it's a negation. When you multiply x^3 by -sin(x), you get -x^2 sin(x), and when you add that, it becomes a subtraction.

That is, what you have is (3x^2)(cos(x)) + (x^3)(-sin(x)) = 3x^2 cos(x) - x^3 sin(x).

As for the tangent line, the slope is the derivative at the given point. Since you want the tangent at x = π, the slope is f'(π) = 3π^2 cos(π) - π^3 sin(π).

#### JeffM

##### Elite Member
You absolutely need to distinguish between a function that is a product of two functions and a composition of 2 functions.

$$\displaystyle u(x) \text { and } v(x) \text { are functions of } x.$$

$$\displaystyle p(x) = u(x) * v(x) \implies \dfrac{dp}{dx} = \dfrac{du}{dx} * v + u * \dfrac{dv}{dx}.$$

That is the product rule.

$$\displaystyle q(x) = u(v(x)) \implies \dfrac{dv}{du} * \dfrac{du}{dx}.$$

That is the chain rule.

Much of the mechanics of differential calculus involves breaking down complex functions into some combination of simpler functions the derivatives of which are determinable by rule.