Find the derivative of h(x) =logbase8(2/x^3)?

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Answer 1 · 2018-04-21T21:04:17

#{d[h(x)]}/dx=-{1}/{xln(2)}#

#h(x)=log_8(2/x^3)#

#{d[h(x)]}/dx=d/dx[log_8(2/x^3)]#

It is worth noting that the derivative of #log_ax# is #1/{xlna}#

Using the chain rule and substitution:

Let #u=2/x^3#

#{d[h(x)]}/dx=d/{du}[log_8(u)] times d/dx[2/x^3]#

#{d[h(x)]}/dx=1/{u ln8} times d/dx[2x^{-3}]#

#{d[h(x)]}/dx=1/{(2/x^3) ln8} times -6x^{-4}#

Simplifying:

#{d[h(x)]}/dx=-{6x^3}/{2 times x^{4} ln8}#

#{d[h(x)]}/dx=-{3}/{xln8}#

#{d[h(x)]}/dx=-{3}/{xln(2^3)}#
#{d[h(x)]}/dx=-{3}/{3xln(2)}#

#{d[h(x)]}/dx=-{1}/{xln(2)}#