Help for calculus exam: partial derivatives

[tarmo]

have a math exam coming up, there's Derivatives if anyone could make me a step by step instruction.
task:

task 2:

task 3:


thanks everyone.

 

[stapel]

have a math exam coming up, there's Derivatives if anyone could make me a step by step instruction.
task:
\(\displaystyle \frac{\delta y}{\delta x}\left((6x^2)(5x^2\, +\, 1)\right)\)

task 2:
\(\displaystyle \frac{\delta y}{\delta x}\left(x\, -\, \ln(x)\right)\)

task 3:
\(\displaystyle \frac{\delta y}{\delta x}\left(\tan(2x)\, -\, \sin(2x)\right)\)

I'm not sure why these are partial derivatives, since there's only apparently the one independent variable...?

You say you are needing "step-by-step instruction". You already have this in your textbook and in your classnotes. In order to help you find where you're getting stuck, it would be better if you replied showing what you've tried so far. Then we can help you get un-stuck.

Please be complete. Thank you!

 

[HallsofIvy]

have a math exam coming up, there's Derivatives if anyone could make me a step by step instruction.
task:
View attachment 2881
task 2:
View attachment 2882
task 3:
View attachment 2883

thanks everyone.

Are you really taking a Calculus class? You don't even seem to know the notation.
"\(\displaystyle \frac{\partial y}{\partial x}((6x^2)(5x^2+ 1))\)"
means "The (partial) derivative of y with respect to x multiplied by \(\displaystyle (6x^2)(5x^2+ 1)\)"
which we cannot do because we do not know what y is.

I think you meant
\(\displaystyle \frac{d(6x^2)(5x^2+ 1)}{dx}\)
The simplest way to do that is to multiply: \(\displaystyle (6x^2)(5x^2+ 1)= 30x^4+ 6x^2\)
and use the fact that the derivative of \(\displaystyle x^n\) is \(\displaystyle nx^{n-1}\).

Similarly for the second and third you need the facts that d(ln(x))/dx= 1/x, \(\displaystyle d(tan(x))/dx= sec^2(x)\), and d(sin(x))/dx= cos(x), together with the "sum", "product", and "chain" rules, all of which you should know.