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### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 16: Sine & cosine derivatives

# Worked example: Derivatives of sin(x) and cos(x)

Dive into the derivative of the function g(x) = 7sin(x) - 3cos(x) - (π/∛x)². By applying the power rule and the derivatives of sine and cosine functions, we efficiently determine the derivative g'(x) = 7cos(x) + 3sin(x) + 2π²/3 * x^(-5/3). Through algebraic manipulation and careful attention to detail, we tackle the problem's initially intimidating appearance.

## Want to join the conversation?

• At - It would be helpful to have more explanation regarding "pi squared, but that's just a number". Why doesn't the last term become 2pi*-2/3x^-5/3?
• When taking derivatives, pi is not taken as a variable. It is instead taken as a constant, so the constant multiple rule can be applied to pi. Hope this helps.
• Why do we need to multiply the exponent to the rest of the terms? why cant we just apply the power rule straight away? i.e. (pi/cube root of x)^2 to 2(pi/cube root of x)^1? I know its the wrong answer? I just don't know why its the wrong answer given I applied to rule correctly? Sorry if its a stupid question but I feel like I'm missing something in my understanding.
• You can actually use the Chain Rule on that term. But you can't use the power rule because inside the parenthesis there is a "nested function" represented by the radical sign. You can only use the power rule (instead of the chain rule) when it's just a simple 'x' that's being differentiated. But that's not the case in this term.
• How would you compute the derivative of -sinx? Do you factor the negative sign so that it's the derivative of -(sinx) which is -cosx?
Also, I don't understand what d/dx means..
• Yes you are correct that the derivative of -sinx is -cosx.
d/dx means "the derivative of, with respect to x". So for example, d/dx (-sinx) = -cosx.
• At ,
Why I can't just write the derivative of the last one putting 2 before it ?
Like 2(pi/cubic square of x)
• That would be invalid since you would be leaving out the expression inside the parentheses unchanged.

As an example - if the derivative of x^6 is 6x^5, what's the derivative of (x^3)^2 then? Is it 2x^3 or x^6 as well? The term in the video is another instance of the same case, just a little bit more complicated.

Anyway, you don't have to fall back to the exponent properties all the time, and you won't be able to as well. But there's another method of dealing with such cases that you might be familiar with already. It's called the chain rule and it starts out exactly as you had suggested and then does the same to the remaining layers of the expression.
• If I were trying to find the derivative of 1/x, what rule would I use to find that?
(1 vote)
• Because 1/x can be rewritten as x^(-1), you can use the power rule to find the derivative of 1/x.
Have a blessed, wonderful day!
• why is the derivative of 9x= 9?
if we calculate using the power rule, shouldnt we be getting the value 9x??
• 9x is 9x¹. By power rule, we find the derivative by multiplying by the exponent (1), then decreasing the exponent by 1.

So we get 1·9x¹⁻¹
9x⁰
9·1
9
• Pi squared is a constant value, so when differentiated its value should be zero right?
• Only if there would be a plus or minus between the 𝝅² and the xˆ-(2/3). If it's bound together by multiplication (or division) you only need to derive one x, so you let 𝝅² live.
• Isn't the derivative of e^x just e^x, and is it because some fundamental property of e?
• The derivative of e^u = e^u*du/dx. Therefore, if u=x, the derivative would equal e^x*1, which is the same as e^x. An example of something more complex, such as the derivative of e^x^2 would be: u=x^2, so the answer would be 2e^x^2.
• In the very last term at the end, when he merges 2/3 and pi^2, how do we know the 2 exponent on the pi applies only to the pi and not the entire 2pi/3?

Did he forget parentheses around the pi^2?
• In any expression with both an exponent and multiplication (like the one you pointed out), the order of operations says we simplify the exponent first (assuming there are no parentheses). So 2pi^2 = 2 * (pi^2). If you wanted to express the square of 2pi you would use parentheses like so: (2pi)^2. Hope this helps! (:
(1 vote)
• shouldn't d(pi)/dy be zero making the part in yellow also all zero?
• I see your confusion. However, you are thinking about taking the derivative of that term in the wrong light.

I guess that you are thinking of finding the derivative of π² * x^(-2/3) by finding the derivative of both π² and x^(-2/3) and multiplying. However, we do not solve for derivatives in that way.

One method is to use the Product Rule to solve the derivative, since there is multiplication in the term. Let's do that.

Formula for the Product Rule
d/dx(u * v) = du/dx * v + u * dv/dx

u = π², v = x^(-2/3) → du/dx = 0, dv/dx = (-2/3)(x^(-5/3))

Product Rule: 0 * x^(-2/3) + π² * (-2/3)(x^(-5/3)) = (-2/3)(π²)(x^(-5/3))

Another method (which is quicker, and can take some practice) is to realize that π² is a constant, and solve for the derivative of x^(-2/3) alone, multiplying the π² back in later.

In this case:
u = x^(-2/3)
du/dx = (-2/3)(x^(-5/3))
π² * du/dx = (π²)(-2/3)(x^(-5/3))

Hope this helps!