# FunctionsDifferentiate the function: f(x)= sinx*lnx

### 3 Answers | Add Yours

You need to use the derivative rules to evaluate the derivative of the function `f(x) = sin x*ln x` , such that:

`(u*v)' = u'*v + u*v'`

Reasoning by analogy yields:

`f'(x) = (sin x*ln x)' => f'(x) = (sin x)'*(ln x) + (sin x)*(ln x)', `

`f'(x) = cos x*ln x + sin x*(1/x)`

**Hence, evaluating the derivative of the function yields **`f'(x) = cos x*ln x + sin x*(1/x).`

f(x)=sinx*Inx

f'(x)=(inx)*(sinx)'+(sinx)*(Inx)'{product rule}

f'(x)=(Inx)*(cosx)+(sinx)*(1/x)

f'(x)=cosxInx+sinx/x

** Therefore the derivative of the function f(x)=sinx*Inx is cosxInx+sinx/x**

(df/dx)*(dg/dx) = d/dx(sinx * lnx)

d/dx(sinx * lnx) = [d/dx(sinx)]*(lnx) + (sin x)*[d/dx(lnx)]

d/dx(sinx * lnx) = cosx*ln x + sin x/x

We've differentiated the product applying the rule:

d(f*g)/dx = (df/dx)*g(x) + (dg/dx)*f(x)

f(x) = sin x

We'll differentiate with respect to x:

df/dx = (cos x)*(x)'

df/dx = cos x

g(x) = ln x

(dg/dx) = (1/x)*(x)'

(dg/dx) = 1