Question

Find an equation of the tangent line to the curve at the given point.

Answer 1

\[y=\sin 5x +\sin^2 5x\]

Answer 2

(0,0)

Answer 3

To find the equation of any line, you'll need to find the slope and one starting point on the line.

Recall that the derivative of a function represents the slope of the tangent line to the curve at any given point (x,y). Can you find the derivative the the given function?

Answer 4

Did you find dy/dx ?

Answer 5

is it 15cos5x?

Answer 6

you forgot about the chain rule for sin^2(5x)

Answer 7

ohh so is it 2cos(5x) *5

Answer 8

do it in separate steps.
first what do you get for sin 5x ?

Answer 9

the 2nd part is more complicated

Answer 10

cos5x

Answer 11

what is the derivative of x^2? that would be 2x

if you replace the 'x' with 'sin(x)' then you'd have 2*sin(x)

then you'd use the chain rule to derive sin(x) to get cos(x)

you would multiply this onto what is already there, so you'd have 2*sin(x)*cos(x)

hopefully this sounds familiar from the lesson?

Answer 12

oh yes

Answer 13

almost. d (sin u) = cos u du
notice the du part

Answer 14

in your case, u = 5x so you need a factor of d (5x)
here d is short for \[ \frac{d}{dx} \]

Answer 15

1?

Answer 16

\[ \frac{d}{dx} \sin 5x = \cos 5x \cdot \frac{d}{dx} 5x\]

Answer 17

so would the d/dx of 5x be 5?
and then i would multiple is by cos5x?

Answer 18

`so would the d/dx of 5x be 5?` correct

`then i would multiple is by cos5x?` also correct

so if,
y = [sin(5x)]^2
then,
dy/dx = 2*sin(5x)*cos(5x)*5 = 10*sin(5x)*cos(5x)

Answer 19

ohhhh

Answer 20

so then in total
y=cos5x +10sin(5x)cos(5x)

Answer 21

`so then in total `
`y=cos5x +10sin(5x)cos(5x)`

close, but not quite

Answer 22

what am i missing?

Answer 23

you differentiated sin(5x) incorrectly on the first term

Answer 24

oh is it 5cos5x

Answer 25

yes, so
If
\[\Large y=\sin(5x) +\sin^2(5x)\]
then
\[\Large \frac{dy}{dx}=5\cos(5x) +10\sin(5x)\cos(5x)\]

Answer 26

oh~

Answer 27

so now I would just plugi n 0 for x to find the slope of the line?

Answer 28

If you want, you can factor out the GCF 5cos(5x)
\[\Large \frac{dy}{dx}=5\cos(5x) +10\sin(5x)\cos(5x)\]
\[\Large \frac{dy}{dx}=5\cos(5x)\left(1 +2\sin(5x)\right)\]

Answer 29

`so now I would just plugi n 0 for x to find the slope of the line?`
yes

Answer 30

5

Answer 31

correct

Answer 32

y=5x

Answer 33

so you know the slope of this tangent line is m = 5 and that this tangent line goes through (0,0)

Answer 34

yep `y = 5x` is the final answer

Answer 35

Thank you so much!! You are so helpful!

Answer 36

you're welcome