# If g(x) = sin(ln x), what is g'(x)

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It is given that g(x) = sin(ln x). The derivative g'(x) has to be determined.

Use the chain rule

g'(x) = [sin(ln x)]'

=> cos (ln x)*[ln(x)]'

=> cos(ln x)*(1/x)

**The required derivative g'(x) = `(cos(ln x))/x` **

We'll use the chain rule to find out the derivative of the function. Let's say that we have two functions, u(v(x)) and v(x). You can see that u is the outside function and v is the inside function.

The chain rule is acting in this way: the derivative of the outside function multiplied by the derivative of inside function.

u(v(x)) = u'(v(x))*v'(x)

let v(x) = ln x and u(v(x)) = sin(ln x)

[sin(ln x)]' = sin'(ln x)*(ln x)'

[sin(ln x)]' = [cos(ln x)]*(1/x)

**Therefore, the derivative of g(x), using the chain rule is g'(x)=[cos(ln x)]/x.**

g(x) = sin(lnx)

Sol: we have to differentiate with respect to x.

g'(x) = d/dx sin(lnx)

by using chain rule

g'(x) = cos(lnx) d/dx lnx

= cos(lnx) . 1/x

g'(x) = cos(lnx)/x