Question

Find the equation of the tangent to the curve
y=x(sqrt 2x^2+7) at x = 3

Answer 1

Do we know how to get the derivative of thisfunction, or is thatthe main problem?

Answer 2

that is the main problem'

Answer 3

do i have to f(x+h)-f(x) / h ??

Answer 4

... Differentiate it using the product rule .

Answer 5

is it x^-1(4x+7)^1/2 ??

Answer 6

Nah. Have you learned about the product rule for derivatives?

Answer 7

the sqrt always stuffs me up

Answer 8

x^n = nx^n-1
??

Answer 9

Well, that is the basic rule. There are also rules for products, quotients, as well as a chain-rule. This problem requires product rule and chain-rule.

Answer 10

any chance you can do the working out to show me??

Answer 11

Well this problem won't require simplifying, merely getting the derivative part correct. So the product rule is this:
\[f'(x)g(x)+f(x)g'(x)\]
That look familiar at all?

Answer 12

yep

Answer 13

Alright. And the chain rule....I don't like describing that as muchwith regular notation. The chain rule is when you have "layers" to a problem. Usually powers, square roots, bunches of grouping symbols, etc. The idea you multiply the derivatives of each layer you have. It's best to show that one with examples rather than word it out, but we'll go through that.

So just to make sure you're doing some thinking, what are our two products for the product rule?

Answer 14

x and 2x^2

Answer 15

Right. So derivative of x?

Answer 16

1x

Answer 17

the first x

Answer 18

2nd x is 4x^1

Answer 19

Just 1. You drop the x basically :3. So the first part of our product rule is f'(x)g(x), so since we got f'(x), Ill fill that in.
\[(1)\sqrt{2x ^{2}+7}+...\]

Answer 20

Well, this is where we gotta know about chain rule and how to deal with derivatives of square roots and such. So you know the formula nx^(n-1). So for a square root, what's the n?

Answer 21

1/n?

Answer 22

Well just 1/2. If you know what I mean.

Answer 23

Now like I said, this is chain rule, meaning there are layers: We have the square root (the 1/2 power) which is an outer layer and then what is inside of the square root is an inner layer. Chain rule says we take the derivative of each layer and multiply them. So the oute rlayer was a square root, meaning its a 1/2 power:
\[(?)^{\frac{ 1 }{ 2 }}\]
Now don't worry about what's inside the root, ignore it. What do I do with that power?

Answer 24

n-1

Answer 25

Where does the 1/2 go?

Answer 26

do we mulitply it to the first x?

Answer 27

We just bring it down in front to multiply. So that means wehave this:
\[\frac{ 1 }{ 2 }(?)^{\frac{ 1 }{ 2 }-1}\]

Answer 28

We bring the power down as a multiplication and then lower the power by 1. So 1/2 - 1 is negative 1/2, meaning we have:
\[\frac{ 1 }{ 2 }(?)^{-\frac{ 1 }{ 2 }}\]

Answer 29

1/2 (2x^2)^1/2 ?

Answer 30

Careful, its negative 1/2 power now.

Answer 31

**-1/2

Answer 32

Right. Now we can worry about the inside. I left it blank because I didnt want us to worry about it. So the inside of the square root was
\[2x ^{2}+7\]
Do you know the derivative of that?

Answer 33

4x^1

Answer 34

Right. So again, chain rule means we are multiplying the derivatives of layers. So our first layer gave us the derivative of:
\[\frac{ 1 }{ 2 }(?)^{-\frac{ 1 }{2 }}\]So now this gets multiplied by our inner derivative of 4x.
\[2x(?)^{-\frac{ 1 }{ 2 }} \]So this was the derivative of the second product. So the first part of the product rule is derivative of the first product multiplied by the 2nd. Now we have the first product multiplied by the derivative of the 2nd. So all of this that we now have gets multiplied x.
Now I left that question mark in there, but we need to put it back. Just know that the inner layer doesnt change, it always stays there. So plugging everything together I have:
\[(2x ^{2}+7)^{\frac{ 1 }{ 2 }}+2x ^{2}(2x ^{2}+7)^{-\frac{ 1 }{ 2 }}\]
So that's our final derivative. We don't need to simplify that really. Now we just plug in x = 3 and get our value. Think you can do that?

Answer 35

yep thanks heaps

Answer 36

Well, you still need an equation of the tangent line, right?

Answer 37

Or you know how to do that part? :P

Answer 38

and the i have my slope of the tangent right?

Answer 39

Yeah, after plugging in x = 3, youll have the slope of your tangent.

Answer 40

thanks heaps how do i give you a medal?

Answer 41

Best response, lol.