You are here:

Advertisement

QUESTION: Greetings,

I look at calculus books and examples on various internet sites and I don't see the "hops" in manipulation of the terms.

I seem to be getting tripped up on problems at the point of recognizing where to parse the expression to apply differential rules. Do you have some method of identifying what to manipulate a given expression to in order to solve it? Can you elaborate on this aspect of "figuring out what to do" when presented a problem. e.g. find the second derivative of y= 3x cos (x^2). Thanks!

ANSWER: The problem is a product rule.

We have y = f(x)g(x), so y' = f'(x)g(x) + f(x)g'(x).

Since f(x) = 3x, f'(x) = 3.

Since g(x) = cos(3x²), the chain rule is used to get g'(x) = -sin(3x²)[6x].

Putting the 6x in front gives -6x*sin(3x²).

Putting these in gives y' = 3cos(3x²) - 18x²sin(3x²).

To find y", the 1st term is done with the chain rule alone and

the 2nd term is done with the product rule and chain rule.

For 3*cos(3x²), the derivative of cos() is -sin() and the derivative of 3x² is 6x.

Putting the minus sign and multiplying the front by 6 gives This gives -18x*sin(3x²).

For -18x²sin(3x²), the product rule says to take f(x) = - 18x² and g(x) = sin(3x²).

This gives f'(x) = -36x and g'(x) = 6x*cos(3x²). That makes the derivative fg' + gf' be

[-18x²][6x*cos(3x²)] + sin(3x²)[-36x].

Multiplying terms together gives -108x³*cos(3x²)] - 36x*sin(3x²).

Note that we could factor out -36x, giving -36x[-3x²*cos(3x²)] + sin(3x²)].

---------- FOLLOW-UP ----------

QUESTION: Thanks for given my question a try Scotto. Given: y= 3x cos (x^2)

F(x) = 3x derivative becomes f'(x)= 3

G(x) is cos x^2 so (cos x^2) becomes - sin x^2 and derivative of x^2 becomes (2x) right?

so I got lost on why you have it = -sin 3x^2 (6x)

I never got to the rest because I got lost here, the rest will probably make perfect sense once I get over this stumbling block.

For some reason I put a 3 in the cos(), and as you pointed out, its not there.

We have y = 3x cos(x²).

This makes y' = 3x * (-2x sin(x²) + 3 cos(x²).

That is, y' = -6x sin(x²) + 3 cos(x²).

This makes y" = -6x * 2x*cos(x²) - 6 sin(x²) - 3*2x*sin(x²).

That works out to y" = -12x²*cos(x²) - 6(1+x)sin(x²).

Right?

Calculus

Answers by Expert:

Any kind of calculus question you want. I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology (reproduction, insusion of chemicals into bloodstream).

Experience in the area: I have tutored students in all areas of mathematics since 1980.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
**Publications**

Maybe not a publication, but I have respond to well oveer 8,500 questions on the PC.
Well over 2,000 of them have been in calculus.
**Education/Credentials**

I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few courses at college a year early.
**Awards and Honors**

I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
**Past/Present Clients**

My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you from all over the world.