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Chirag, Master's Degree
Category: Calculus and Above
Satisfied Customers: 11954
Experience:  I have a Master's degree in Engineering and a very wide teaching experience of more than 25 years at various levels.
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# f(X)=（x³+2x）e^x ； y=e^ax³； y=e^-2t cos4t； f(t)=sin(e^t)+e^sint; y=e^u-e^-u/e^u+e^-u;

### Resolved Question:

f(X)=（x³+2x）e^x ；
y=e^ax³；
y=e^-2t cos4t；
f(t)=sin(e^t)+e^sint;
y=e^u-e^-u/e^u+e^-u;
y=√1+xe^-2x
Differentiate the function.
Sorry i am not good at math.
Submitted: 1 year ago.
Category: Calculus and Above
Expert:  Chirag replied 1 year ago.

Welcome and Thanks for requesting me.

There are 6 problems in all right? Need only answers? Let me know.

Thanks.

Customer: replied 1 year ago.
NO,I need every answer for every problem. Thank you. I am not good at math.
Expert:  Chirag replied 1 year ago.
You need working steps for each problem? Or only the final answers for each?
Customer: replied 1 year ago.

y=e^-2t cos4t；
f(t)=sin(e^t)+e^sint;
y=e^u-e^-u/e^u+e^-u;
y=√1+xe^-2x;

I need those questions answer , if you have step for each problem that is better. Thank you. Just four questions.

Expert:  Chirag replied 1 year ago.
Ok, I am working on your questions and will post the answers shortly.

Expert:  Chirag replied 1 year ago.

(1) f'(x) = (x^3 + 2x) e^x + (e^x)(3x^2 + 2)

= (e^x) (x^3 + 3x^2 + 2x + 2)

(2) y = (e^(ax^3)) a(3x^2)

= (3ax^2) e^(ax^3)

(3) y' = (e^-2t)(-4 sin 4t) + (cos 4t)(-2 e^-2t)

= (e^-2t)(-4 sin 4t - 2 cos 4t)

(4) f'(t) = (cos e^t) e^t + (e^(sin t)) cos t

(5) y' = [(e^u + e^-u)(e^u + e^-u) - (e^u - e^-u)(e^u - e^-u)] / (e^u + e^-u)^2

= 4 /(e^u + e^-u)^2

(6) y' = [1/2 √(1 + xe^-2x)] [x(-2e^-2x) + (e^-2x)]

= (e^-2x)(1 - 2x) / [2 √(1 + xe^-2x)]

I hope that helps. Let me know if you need any clarifications on my answers. I will be happy to provide them. Else, please give a rating for my answer.

Thanks.

Chirag, Master's Degree
Satisfied Customers: 11954
Experience: I have a Master's degree in Engineering and a very wide teaching experience of more than 25 years at various levels.

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