Calculus: Finding First Derivatives for Functions - Homework Help

In summary, the conversation is about someone asking for help with their homework and providing their attempts at solving the problems. They are seeking clarification and confirmation on their solutions and are open to receiving assistance. The conversation also mentions the use of LaTeX for easier readability. A link is provided for the use of LaTeX in posts.
  • #1
DTskkaii
19
0
Hi, I was hoping someone could help me out with my homework set.
I have done a lot of the questions, and it would help if someone could tell me if I have done them correctly. Thanks! :)

Q1: Find first derivatives for the following functions
(a)g(s,t)=sin(st^3)
g_t=δg/δt=δ/δt*(sin(st^3))=t^3cos(st^3)
g_s=δg/δs=δ/δs*(sin(st^3))=3st^2cos(st^3)
Just wrote that out full for working, will shorten now
(b) f(x,y)=x(y^3)+(2x^4)y
f_x=(y^3)+8(x^3)y
f_y=3x(y^2)+2(x^4)
(c) g(r,x,z)=rsin(zx)
g_r=sin(zx)
g_x=rzcos(zx)
g_z=rxcos(zx)
(d) e_(X1,X2...Xn)=sqrt(X1^2+X2^2...Xn^2)
only write down one generalised partial derivative with respect to Xi
I am not sure how to approach this one, help would be great
I also am not entirely sure if I am somehow meant to put the two partial derivatives back together to get the complete first derivative?

Q2: y(x,t)=Asin(kx-wt) where w=(pi/2), k=pi, A=5
(a) find rate of change of y wrt to t at x=1, t=1
δy/δt=-Awcos(kx-wt)
=-(5/2)pi*cos(pi/2)
=-7.851
(b) find rate of change of y wrt to x at x=(1/2), t=1
δy/δx=-Akcos(kx-wt)
=-5pi*cos0
=15.708

If anyone could tell me if I am on the right track with these questions, and help out with Q1 (d), I would be super appreciative.
 
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  • #2
DTskkaii said:
Hi, I was hoping someone could help me out with my homework set.
I have done a lot of the questions, and it would help if someone could tell me if I have done them correctly. Thanks! :)

Q1: Find first derivatives for the following functions
(a)g(s,t)=sin(st^3)
g_t=δg/δt=δ/δt*(sin(st^3))=t^3cos(st^3)
g_s=δg/δs=δ/δs*(sin(st^3))=3st^2cos(st^3)
Just wrote that out full for working, will shorten now
(b) f(x,y)=x(y^3)+(2x^4)y
f_x=(y^3)+8(x^3)y
f_y=3x(y^2)+2(x^4)
(c) g(r,x,z)=rsin(zx)
g_r=sin(zx)
g_x=rzcos(zx)
g_z=rxcos(zx)
(d) e_(X1,X2...Xn)=sqrt(X1^2+X2^2...Xn^2)
only write down one generalised partial derivative with respect to Xi
I am not sure how to approach this one, help would be great
I also am not entirely sure if I am somehow meant to put the two partial derivatives back together to get the complete first derivative?

Q2: y(x,t)=Asin(kx-wt) where w=(pi/2), k=pi, A=5
(a) find rate of change of y wrt to t at x=1, t=1
δy/δt=-Awcos(kx-wt)
=-(5/2)pi*cos(pi/2)
=-7.851
(b) find rate of change of y wrt to x at x=(1/2), t=1
δy/δx=-Akcos(kx-wt)
=-5pi*cos0
=15.708

If anyone could tell me if I am on the right track with these questions, and help out with Q1 (d), I would be super appreciative.

Your computation of g_t in (a) is incorrect; you need to use the Chain Rule to get a proper computation. You should have gotten [itex]g_t = 3 s t^2 \cos(st^3).[/itex] I did not check the others, so there may or may not be additional errors.

RGV
 
  • #4
micromass said:
DTskkaii,

It may be nice to use LaTeX in your post. That would make it so much more readable and easy for us!

https://www.physicsforums.com/showthread.php?t=546968

I'm glad you posted this link, as i just made it into my signature. Hopefully, it will help spread the message. :smile:
 

What is a derivative in calculus?

A derivative in calculus is a mathematical concept that represents the instantaneous rate of change of a function at a specific point. It can also be thought of as the slope of a tangent line to the graph of the function at that point.

What is the process of finding a derivative?

The process of finding a derivative is known as differentiation. It involves using specific rules and formulas to calculate the slope of a tangent line at a given point. These rules include the power rule, product rule, quotient rule, and chain rule.

What is the purpose of derivatives in calculus?

The purpose of derivatives in calculus is to help us understand and analyze the behavior of functions. They can be used to find maximum and minimum points, determine the rate of change of a function, and solve optimization problems.

What is the difference between a derivative and an antiderivative?

A derivative and an antiderivative are inverse operations of each other. A derivative represents the instantaneous rate of change of a function, while an antiderivative represents the original function before differentiation.

How are derivatives used in real-world applications?

Derivatives have many real-world applications, such as in physics, engineering, economics, and biology. They can be used to solve problems involving motion, optimization, growth and decay, and many other real-life scenarios.

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