Question #dcb24

1 Answer
John D.
Oct 7, 2017

#d/dx((2^xcotx)/sqrt2) = (2^x (cotx(ln2)-csc^2x))/sqrt2#

Explanation:

Let's use the product rule:

#d/dx(ab) = b(da)/dx + a(db)/dx#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#d/dx((2^xcotx)/sqrt2) = 1/sqrt2 d/dx(2^xcotx)#

Let's ignore the #1/sqrt2# for now; we can put it back once we're done.

#d/dx(2^xcotx) = 2^x(d/dxcotx) + cotx(d/dx2^x)#

#= 2^x(-csc^2x)+cotx(2^xln2)#

#= -2^xcsc^2x + 2^xcotx(ln2)#

#= 2^x (cotx(ln2)-csc^2x)#

Now that we have our derivative, let's divide it by the #sqrt2# we removed in the beginning.

#d/dx((2^xcotx)/sqrt2) = (2^x (cotx(ln2)-csc^2x))/sqrt2#

Final Answer

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