How Do You Calculate Derivatives for Trigonometric and Exponential Functions?

[BuBbLeS01]

The first problem is...
y = 1/2 e^x - 3 sinx
Derivative = 1/2e^x - 3 cosx * 1
Is that right? I wasn't sure about the 1/2.

The second problem I have is...
f(x) = (2e^x)/(x^2 + 1)
I am not sure what the derivative of 2e^x is...is it just 2e^x?
(x^2 + 1)(2e^x) - (2e^x)(2x)/(x^2 + 1)^2

We don't have to simplify.

 

[bob1182006]

the first one's right.

and so is the second one.

if you have 2e^x the 2 is just a constant so it's the same as 2*d(e^x)/dx

 

[rocomath]

We don't have to simplify.

:O even on the test? either way it's good practice to simplify.

 

[BuBbLeS01]

Thank you guys. No he doesn't want us to simplify on tests either because it is easier to grade for him...hehe.

 

[BuBbLeS01]

So for 2 sinxcosx the derivative would be...
(2cosxcosx)*(-sinx)

 

[BuBbLeS01]

And for x^2 tanx^2 you would treat that as 3 separate derivatives?
(2x) * (sec^2x^2) * (2x)

 

[rocomath]

So for 2 sinxcosx the derivative would be...
(2cosxcosx)*(-sinx)

what trig identity could you use?

2sinxcosx=? by doing that, your derivative becomes much easier.

 

[rocomath]

And for x^2 tanx^2 you would treat that as 3 separate derivatives?
(2x) * (sec^2x^2) * (2x)

3? i only see the product rule

$$x^{2}\tan^2{x}$$

correct?

 

[BuBbLeS01]

what trig identity could you use?

2sinxcosx=? by doing that, your derivative becomes much easier.

Trig identity is sin2x right?
(cos2x) * (2)

 

[BuBbLeS01]

3? i only see the product rule

$$x^{2}\tan^2{x}$$

correct?

Yea that's right. I thought you took the derivative of x^2 then tan x^2 and x^2

 

[rocomath]

Yea that's right. I thought you took the derivative of x^2 then tan x^2 and x^2

i think you mean you will need to do the chain rule

if so, that is correct

 

[BuBbLeS01]

i think you mean you will need to do the chain rule

if so, that is correct

Oh yes the chain rule, sorry. So this is correct...
(2x) * (sec^2x^2) * (2x)

 

[rocomath]

Oh yes the chain rule, sorry. So this is correct...
(2x) * (sec^2x^2) * (2x)

no! what book are you using?

$$x^{2}\tan^{2}x$$

can be re-written as

$$x^{2}(\tan{x})^2$$

how do you differentiate and apply the chain rule to tanx?

 

[BuBbLeS01]

Oh okay forgot about that...
2x(tanx)^2 * (sec^2x)^2
do you also multiply by 2(tanx) or no.

 

[rocomath]

Oh okay forgot about that...
2x(tanx)^2 * (sec^2x)^2
do you also multiply by 2(tanx) or no.

differentiating and applying the chain rule to (tanx)^2

$$2\tan{x}\sec^2{x}$$

 

[BuBbLeS01]

Okay so the final answer would be...
(2x(tanx)^2) * 2(tanx sec^2x)

 

[rocomath]

Okay so the final answer would be...
(2x(tanx)^2) * 2(tanx sec^2x)

it shouldn't be times, it should be + ? (product rule)

 

[BuBbLeS01]

I thought you said to use the chain rule?

 

[rocomath]

I thought you said to use the chain rule?

were you giving me the answer to x^2tan^2x or just tan^2x?

if you were giving the overall answer, then it's still incorrect.

 

[BuBbLeS01]

the tan x^2
overall answer being...
(2x) + 2(tanx sec^2x)

 

[BuBbLeS01]

is that right now?

 

[BuBbLeS01]

can someone let me know if that is right?

 

[rocomath]

the tan x^2
overall answer being...
(2x) + 2(tanx sec^2x)

no it is still wrong.

 

[BuBbLeS01]

I don't understand then what I am doing wrong.

 

[rock.freak667]

$$y=x^2tan^2x = x^2(tanx)^2 = (xtanx)^2$$

$$let u = xtanx$$ then use the product rule to find $$\frac{du}{dx}$$ and then use the chain rule

 

[BuBbLeS01]

so (xsec^2x + tanx) * 2(xtanx)

 

[rock.freak667]

so (xsec^2x + tanx) * 2(xtanx)

that would appear to be correct from what I can see

 

[rocomath]

so (xsec^2x + tanx) * 2(xtanx)

yep, since your teacher doesn't want you to simplify further