Calculus 6.3b - Additional Examples

Some more examples dealing with differentiating natural logarithms.

Category:

Education

Tags:

• Calculus
• logarithms
• exponents
• distance learning
• homeschool
• Derek Owens

[brgn9099]

Why is ex 7 not a product rule? Isnt it the same as cos x ln2x?

[derekowens]

@brgn9099 Yes, I think it could be done that way as well.

[ImMoreBetter]

Thanks for the video(s). They really helped me. Test tomorrow... >_>

Note to self: Don't fall asleep in calculus again!

[ChavezMath]

@derekowens I would of answered, NO, brgn9099 introduced the variable x (cos x) that wasn't there to begin with. As you mention in the video this is a chain rule problem.

d/dx(cos x ln 2x) = -sin x ln 2x + cos x *1/x = -sin x ln 2x + (cos x) / x.

This result is completely different from your result in the video.

[ckala13]

@imwithdumbhead You can't use the product rule here because the product rule is only used to find the derivative of two functions that are being multiplied by each other. Example 7 involves taking the derivative of a composition of functions, not the derivative of a product of two functions. d/dx f(g(x)) = f '(g(x)) * g'(x). d/dx f(x)g(x) = f '(x)g(x) + g '(x)f(x).

[ckala13]

@brgn9099 Hi there, this is a late response, but cosxln2x is not the same as cos(ln2x). You wouldn't use the product rule in example 7 because it's a composition of 2 functions, not the product of two functions. If you were to take the derivative of cosxln2x, you should get -sinxln2x +(cosx)/x. (I'm pretty sure I did that right). When you take the derivative of cos(ln2x) however, you get (-sin(ln2x))/x. The two are not equivalent. Hope this helps..

[imwithdumbhead]

@derekowens when you do solve #7 by product rule, it gives cos/x-sin(ln2x). the only difference is the cosine. y?!

[6BURG9]

@brgn9099

i dont think you can do it as a product rule because ln(2x) is a function of cosine.

cos(x)ln(2x) =/= cos(2x)

[wizardman201]

you're the bomb