Question

What is the derivative of 5xsecx+x^3cosx+17(pi)?

Answer 1

You will need to know the derivative of a constant, the derivative of cos(x),the derivative of sec(x), and the derivative of x, the product rule, and the power rule.

So let's look at each term individually:

5xsec(x)


x^3


cos(x)


17pi

Answer 2

5xsec(x) is a product
[(5x) * (sec(x))]'=?

You will need to know the product rule. Hint here: (sec(x))'=tan(x)sec(x)

---
[x^3]'=?

You will need to know the power rule.
----

(cos(x))'=-sin(x)

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17pi is a constant so (17pi)'=?
What does the constant rule say?

Answer 3

[x^3]'= \[3x ^{2}\]

Answer 4

yes

Answer 5

will the derivative of 17pi be 0?

Answer 6

yep

Answer 7

So the last part to look at is (5xsec(x))'

Answer 8

(5x)'=?

Answer 9

Also the product rule is (u*v)'=u'*v+u*v'

Answer 10

(5x)'= x ?

Answer 11

That is like saying the derivative of x is x which is not.

Answer 12

The slope of y=5x is 5 so the derivative of (5x) is 5.

Answer 13

ooh okay so (5xsec(x))' = 5tan(x)sec(x) ?

Answer 14

Well that is a product you need to use the product rule not a madeup one.

Answer 15

(5xsec(x))'
=
(5x)'sec(x)+5x(sec(x))' by product rule

Answer 16

what about tan(x) ? All of my answer choices have tan in them?

Answer 17

Right which is a good thing because our answer as tan in it.

Answer 18

has*

Answer 19

based on what we have said I am leading towards answer choice C which is 5sec(x)tan(x)+3x^2sin(x) is that the answer?

Answer 20

So...well again (5xsec(x))'
is equal to (5x)'sec(x)+5x(sec(x))'

Answer 21

You know (5x)' and (sec(x))'
Just replace them in this:
(5x)'sec(x)+5x(sec(x))'

Answer 22

so 5secx+5xsecxtanx?

Answer 23

cause my other answer choice is 5secx+5xsecxtanx+3x^2cosx-x^3sinx , but I am not sure how they got x^3sinx

Answer 24

What is the derivative of 5xsecx+x^3 + cosx+17(pi)

this was the question right?
this is what you wrote

Answer 25

You didn't write x^3cos(x)
you wrote x^3+cos(x)

Answer 26

:O I meant x^3cos(x)

Answer 27

Use the product rule again since we have a product.

Answer 28

i think the answer is 5sec(x)tan(x)+3x^2sin(x)

Answer 29

What about the other parts?

Answer 30

We just said (5xsec(x))'=5sec(x)+5xsec(x)tan(x)

Now how do we find the derivative of x^3cos(x)?

Answer 31

well 17pi =0 right?

Answer 32

(17pi)'=0

Answer 33

We need to find (x^3cos(x))'
Again we need to use the product rule
(x^3cos(x))'
=
(x^3)'cos(x)+x^3(cos(x))'

Answer 34

=-3x^2sinx

Answer 35

nope there are two terms

Answer 36

yes well that 2 times: -3x^2sinx+3x^2sinx

Answer 37

no no...

Answer 38

(x^3)'=3x^2
(cos(x))'=-sin(x)

(x^3cos(x))'
WE are finding the derivative of a product so we must use the product rule
(x^3)'cos(x)+x^3(cos(x))'
Now looking above
3x^2cos(x)+x^3(-sin(x))

Answer 39

so 3x^2cosx-x^3sinx?

Answer 40

for that part yes

Answer 41

so the answer is 5secx+5xsecxtanx+3x^2cosx-x^3sinx

Answer 42

yes

Answer 43

thank you!

Answer 44

np
I hope you understand the product rule now