What is the derivative of:
a) cos(3x)
b) π/x
I did (a) just by following chain rule:
-sin(3x)*3
so it's -3sin(3x)
Is that correct?
I'm not completely sure how to do (b). Would it just be like the derivative of 1/x? That's -1/(x^2), would it be -π/(x^2) ?
Both of your answers are correct. Nice job!
Yes, you are correct for part (a). The derivative of cos(3x) using the chain rule is -3sin(3x), because the derivative of cos(u) is -sin(u), and you need to multiply it by the derivative of the inner function (in this case, 3x) with respect to x, which is 3.
For part (b), you are on the right track. The derivative of 1/x is indeed -1/(x^2). However, since there's a constant π multiplied to 1/x, you need to consider that as well when taking the derivative.
To find the derivative of π/x, you can use the quotient rule, which states that if you have a function of the form f(x)/g(x), where f(x) and g(x) are both differentiable functions, the derivative is given by [(g(x)*f'(x)) - (f(x)*g'(x))] / [g(x)^2].
In this case, f(x) is the constant π, and g(x) is x. Therefore, f'(x) is 0 (since the derivative of a constant is always 0), and g'(x) is 1.
Applying the quotient rule, we have [(x*0) - (π*1)] / (x^2), which simplifies to -π/(x^2). So, your answer is indeed -π/(x^2).
Well done on both questions!