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Differentiation involving Sin(x) as a power

  • Thread starter Bananarama
  • Start date
  • #1

Homework Statement



Find the derivative of y=(x^2)^sinx; using the chain rule.


Homework Equations



No other relevant equations.


The Attempt at a Solution



I attempted to apply the Chain rule: dy/dx = dy/du X du/dx
Subbing u for x^2, which made y = u^sinx

I ended up with du/dx = 2x;

and dy/du = ln(u)*cos(x)*(U)^sinx

I know this isn't right, and all i've spent the past hour doing is confusing myself. I've searched my text book, and online, and can't find anyting (that I understand) to help me.
To be honest, i'm not even sure how i'm supposed to apply the chain rule

Any help would be appreciated, thank you.
 

Answers and Replies

  • #2
Mute
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The problem with the way you have attempted it is that when you computed dy/du you did not write y entirely in terms of u, you left it in terms of u and x, but u and x are not independent. This is why your calculation isn't working you. You should replace x with ##\sqrt{u}## to do it properly.

There is a slightly easier way to do this, as well. Note that you can write

##x^2 = e^{\ln (x^2)} = e^{2\ln |x|}.##

Then, recall that ##(a^b)^c = a^{bc}## for real numbers a, b and c. This will give you a form that is easier to differentiate. (You'll want to use a different chain rule substitution, though).
 
  • #3
SammyS
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Homework Statement



Find the derivative of y=(x^2)^sinx; using the chain rule.

Homework Equations



No other relevant equations.

The Attempt at a Solution



I attempted to apply the Chain rule: dy/dx = dy/du X du/dx
Subbing u for x^2, which made y = u^sinx

I ended up with du/dx = 2x;

and dy/du = ln(u)*cos(x)*(U)^sinx

I know this isn't right, and all i've spent the past hour doing is confusing myself. I've searched my text book, and online, and can't find anyting (that I understand) to help me.
To be honest, I'm not even sure how I'm supposed to apply the chain rule

Any help would be appreciated, thank you.
I can see a couple of ways to do this with the chain rule. One uses the multi-variable chain rule. The other require rewrite the function.

Multi-variable chain rule:
Let u(x)= x2 and v(x) = sin(x)

Then [itex]y=f(x,\,y)=u^v\,.[/itex]

Then use [itex]\displaystyle \frac{dy}{dx}=\frac{\partial y}{\partial u}\frac{du}{dx}+\frac{\partial y}{\partial v}\frac{dv}{dx}\ .[/itex]

The single variable method:
[itex]\displaystyle y=(x^2)^{\sin(x)}=e^{2\ln(x)\sin(x)}\ .[/itex]

Now apply the chain rule. You will also need the product rule.​
 
  • #4
HallsofIvy
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Homework Helper
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Simpler, I think, than writing [itex]f(x)^{g(x)}[/itex] as [itex]e^{g(x)ln(f(x))}[/itex] is to start with [itex]y(x)= f(x)^{g(x)}[/itex] and take the logarithm of both sides: ln(y)= g(x)ln(f(x)).

Here, the function is [itex]y(x)= (x^2)^{sin(x)}= x^{2sin(x)}[/itex]. Then [itex]ln(y)= 2sin(x) ln(x)[/itex]. Then you can use the product rule to differentiate the right side using the product rule. Of course, the derivative of ln(y), on the left, is (1/y)dy/dx. After taking the derivative of both sides multiply through by [itex]y= (x^2)^{sin(x)}[/itex] to find dy/dx.
 

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