Differentiation Question Concerning natural log

AI Thread Summary
To find the derivative of y = e^(3ln(x^2)), the calculation leads to dy/dx = 6/x * e^(3ln(x^2)). This simplifies to dy/dx = 6x^5 after recognizing that e^(3ln(x^2)) equals x^6. The discussion emphasizes the importance of applying logarithmic rules when encountering expressions involving e and ln. Participants confirm the correct derivative and acknowledge the simplification process. Understanding these relationships is crucial for solving similar differentiation problems.
courtrigrad
Messages
1,236
Reaction score
2
Hello all

If y = e^3^\ln^(x^2) find \frac {dy}{dx}

So \frac {dy}{dx} =(3 (\frac {1}{x^2}) \* 2x e^3^\ln^(x^2)

So the simplified answer is: \frac {6}{x} e^3^\ln^(x^2)

Is this correct? IS there any other way of expressing the answer?

Thanks
 
Physics news on Phys.org
Use your log rules.

e^{3 \ln x^2} \equiv e^{\ln x^6} \equiv x^6

- Warren
 
oh i see

So derivative would be 6x^5

didn't catch that

thanks
 
Pretty much anytime you see e and ln stuck together like that, it means your teacher is trying to get you to use log rules.

- Warren
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...

Similar threads

Replies
10
Views
852
Replies
3
Views
700
Replies
28
Views
1K
Replies
2
Views
721
Replies
4
Views
2K
Replies
44
Views
4K
Replies
6
Views
758
Back
Top