Help with finding first level derivative of y = x^2 * 2^x^2

[9dipity]

Hi there, we're currently doing derivatives at school and Im a bit stuck with one of the homeworks.
Im not sure if Im applying the right rules to the equation.

We have to find the first level derivative.
The equation given is:
y = x^2 * 2^x^2

Hope that makes sense?
Its x to the power of 2 times 2 to the power of x to the power of 2. If that helps?


So the way I understood it, I have to apply certain rules to the equation to gain the result.
In this case - (x^n) = nx^n-1

So my way of thinking:

y = x^2 * 2^x^2 becomes
2x * 2 * 2x

Is this right? Or completely wrong (more likely).

 

[tkhunny]

Hi there, we're currently doing derivatives at school and Im a bit stuck with one of the homeworks.
Im not sure if Im applying the right rules to the equation.

We have to find the first level derivative.
The equation given is:
y = x^2 * 2^x^2

Hope that makes sense?
Its x to the power of 2 times 2 to the power of x to the power of 2. If that helps?


So the way I understood it, I have to apply certain rules to the equation to gain the result.
In this case - (x^n) = nx^n-1

So my way of thinking:

y = x^2 * 2^x^2 becomes
2x * 2 * 2x

Is this right? Or completely wrong (more likely).

Just to be clear, we're talking about this...? \(\displaystyle y = 2^{x}\cdot 2^{x^{2}}\)

If so, this will take a whole lot more effort. Have you met:
* Multiplication Rule
* Logarithmic Differentiation

 

[pka]

Hi there, we're currently doing derivatives at school and Im a bit stuck with one of the homeworks.
Im not sure if Im applying the right rules to the equation.
We have to find the first level derivative.
The equation given is: y = x^2 * 2^x^2 Its x to the power of 2 times 2 to the power of x to the power of 2. If that helps?

If the question is \(\displaystyle D_x\left[(x^2)(2^{x^2}\right]\) then all I can say if that is too advanced question for a beginner.
First of all the question requires logarithmic differentiation.
If \(\displaystyle f(x)\) is a differentiable function then \(\displaystyle D_x\left[2^{f(x)}\right]=\left[2^{f(x)}\right]\cdot f'(x)\cdot\log(2)\)
Now you must apply the above using the product rule as well as the chain rule.
Post your work and it will be reviewed.

 

[9dipity]

I think I got it??

Yeah, sorry, I think I might have messed up my thread posting. First time here.

I had a go at the problem, want to check if I did it right or completely off track


I uploaded to Imgur to make it easier to see rather than typing it out.

Hope someone can correct me if I've done it wrong

Thank you

 

[tkhunny]

Too bad we don't much like imgur links, in here.
No sure how to help if we can't see the actual problem statement.

 

[9dipity]

Too bad we don't much like imgur links, in here.
No sure how to help if we can't see the actual problem statement.

Hope attaching it directly works? hah
Let me know how wrong I am, just be gentle :lol:

 

[9dipity]

Too bad we don't much like imgur links, in here.
No sure how to help if we can't see the actual problem statement.

Any post I do seems to take 24h+ to actually post..
Sorry about this.
Im gonna try attaching an actual photo this time:


Let me know how far off am I? hah

 

[HallsofIvy]

("2^x^2" is ambiguous. It could be 2^(x^2) or (2^x)^2. But since (2^x)^2 would be more easily written 2^(2x) I will assume you mean 2^(x^2).)

Given that \(\displaystyle f(x)= x^22^{x^2}\), let \(\displaystyle y= x^2\). Then f(y)= y2^y. The derivative of \(\displaystyle 2^y\), with respect to y, is \(\displaystyle ln(2)2^y\). (Did you know that? See below). Using the "product rule" then \(\displaystyle \frac{df}{y}= 2^y+ ln(2)y2^y= (1+ ln(2)y)2^y\). Since \(\displaystyle \frac{dy}{dx}= \frac{dx^2}{dx}= 2x\), by the "chain rule", \(\displaystyle \frac{df}{dx}= 2x(1+ ln(2)x^2)2^{x^2}\).

(Let \(\displaystyle f(x)= a^x\). Then \(\displaystyle f(x+ h)= a^{x+ h}= a^xa^h\). The "difference quotient" is \(\displaystyle \frac{f(x+h)- f(x)}{h}= \frac{a^xa^h- a^x}{h}= \frac{a^x(a^h- 1)}{h}= \frac{a^h- 1}{h}a^x\). Taking the limit as h goes to 0, \(\displaystyle \frac{df}{dx}= \lim_{h\to 0}\frac{a^h- 1}{h}a^x\).

That is, the derivative of \(\displaystyle a^x\) is \(\displaystyle a^x\) times the "constant" (depending upon a but not x) \(\displaystyle C_a= \lim_{h\to 0}\frac{a^h- 1}{h}\). It is fairly easy to estimate that constant for various a by taking h small.

For example, if a= 2, taking h= 0.001, we have \(\displaystyle C_2\approx \frac{2^{0.001}- 1}{0.001}= \frac{1.0007- 1}{0.001}= 0.7\). For a= 3, that is \(\displaystyle C_3\approx \frac{3^{0.001}- 1}{0.001}= \frac{1.0011- 1}{0.001}= 1.100\).

Since that constant, \(\displaystyle C_a\), is less than 1 for a= 2 and larger than 1 for a= 3, there must exist a value of a between 2 and 3 such that \(\displaystyle C_a= 1\). We define "e" to be that value of a: \(\displaystyle \lim_{h\to 0}\frac{e^h- 1}{h}= 1\). That is, \(\displaystyle \frac{de^x}{dx}= e^x\).

Now, define "\(\displaystyle ln(x)\)" to be the inverse function to \(\displaystyle e^x\). Then, for general a, we can write \(\displaystyle a^x= e^{ln(a^x)}= e^{xln(a)}\) and, by the chain rule, \(\displaystyle \frac{da^x}{dx}= e^{ln(a^x)}\frac{d(xln(a)}{dx}= ln(a)a^x\).)

 

[Dr.Peterson]

Any post I do seems to take 24h+ to actually post..
Sorry about this.
Im gonna try attaching an actual photo this time:
View attachment 10707

Let me know how far off am I? hah

The short answer is, you're correct. Good work!

How much you simplify the answer is a matter of taste, so some of your work might be considered unnecessary, but none is wrong.