# Differentiate f(x) = sinx * lnx

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### 3 Answers

f(x) = sinx* lnx

To find the first derivative, we will use the product rule:

Let f(x) = u*v such that:

u= sinx ==> u' = cosx

v= lnx ==> v' = 1/x

Now we know that:

if f(x) = u*v

Them f'(x) = u'v + uv'

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

==>** f'(x) = cosx*lnx + sinx/x**

** = (x*cosx*lnx + sinx )/x**

d(sinx.lnx)/dx

= sinx.d(lnx)/dx + lnx.d(sinx)/dx

= (sinx)/x + lnx.cosx

(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