Learn to Find Derivatives: f(x) = 3cosx - (sinx/4) | Quotient Rule Explained

  • Thread starter Thread starter physics=world
  • Start date Start date
  • Tags Tags
    Derivative
physics=world
Messages
109
Reaction score
0
1. f(x) = 3cosx - ((sinx)/(4))

Homework Equations


use the rules of differentiation3. i tried adding them together which i got

((12cosx - sinx) / (4))

then, i used quotient rule

((4)(-12sinx - cosx) - (12cosx - sinx)(0))/ (4^2)

which i got this for an answer

(-12sinx - cosx) /( 4)but the answer is
(-3sinx - cosx) /( 4)
 
Last edited by a moderator:
Physics news on Phys.org
physics=world said:
1. f(x) = 3cosx - ((sinx)/(4))



Homework Equations


use the rules of differentiation


3. i tried adding them together which i got

((12cosx - sinx) / (4))

then, i used quotient rule

((4)(-12sinx - cosx) - (12cosx - sinx)(0))/ (4^2)

which i got this for an answer

(-12sinx - cosx) /( 4)


but the answer is
(-3sinx - cosx) /( 4)
Neither of these is correct, assuming that the problem is as you posted.

If you are learning the quotient rule, you should also have learned some simpler rules, such as the sum rule and the constant multiple rule.

As a matter of course, you should never use the quotient rule if the denominator is a constant. It's not wrong to do so, but since the quotient rule is a bit on the complicated side, there is a greater chance of getting the derivative wrong due to an error of some sort.
 
Mark44 said:
Neither of these is correct, assuming that the problem is as you posted.

If you are learning the quotient rule, you should also have learned some simpler rules, such as the sum rule and the constant multiple rule.

As a matter of course, you should never use the quotient rule if the denominator is a constant. It's not wrong to do so, but since the quotient rule is a bit on the complicated side, there is a greater chance of getting the derivative wrong due to an error of some sort.

ohh. thanks i see what i did wrong.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top