Find the derivative of the following function

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The derivative of the function y = cos^3(5x^2 - 6) requires the application of the chain rule twice, treating both the cosine function and the exponent as separate functions. The correct approach involves first differentiating the outer function, resulting in 3(cos(5x^2 - 6))^2, and then multiplying by the derivative of the inner function, which is -sin(5x^2 - 6) multiplied by the derivative of (5x^2 - 6). The final derivative is -30x cos^2(5x^2 - 6) sin(5x^2 - 6). This confirms that the initial confusion about applying the chain rule was resolved, leading to the correct answer.
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Homework Statement



y = cos^3(5x^2-6)


Homework Equations





The Attempt at a Solution



I used the chain rule for this question but I am not sure about the cos^3 part. Using the chain rule i got

-10xSin^3(5x^2-6)

But was I supposed to use the power rule on the cos^3 part? I can think of 3 ways this could be done but I am not sure which way to go about it.

Thanks!
 
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You have to think of both cosine AND the exponent 3 as external functions. So you need to perform the chain rule twice:

(d/dx) cos^3(5x^2-6) = 3 (cos(5x^2-6))^2 * (d/dx) cos(5x^2-6), at which you point you'll also need to do the chain rule cos(5x^2-6).
 
How do I do the chain rule twice with this?

I got 3 Cos(5x-6)^2 by -sin (5x^2-6)?

Do I do something like ( 3 (-sin (5x^2-6) d/dx (5x^2-6))^2 ) (-sin (5x^2-6)

?
 
I think I was making this too complicated.

=d/dx Cos^3 d/dx (5x^2-6)
=3Cos^2(5x^2-6) -sin(5x^2-6) 10x
=-30xCos^2(5x^2-6)Sin(5x^2-6)

What do you guys an gals think?
 
That is correct.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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