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

1 Answer
Apr 21, 2018

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

Explanation:

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)}