Find differentiation of y = xsin(xlogx) ?

Question

Solve it using chain rule step by step.

Answer 1

I'm assuming that you are writing log in place of ln (log is natural log, not log base 10).

 

Following this assumption, we have 2 differentiation rules which are useful.

 

(1)  Product rule:  If f(x) = g(x)*h(x), then f'(x) = g(x)*h'(x) + g'(x)*h(x)

(2)  Chain rule.

 

Using (1),

y' = x * [derivative of sin(x*ln x)] + (derivative of x) * sin(x*ln x)

   = x * [derivative of sin(x * ln x)] + 1* sin(x * ln x)

 

Using the chain rule to find the derivative of sin(x * ln x),

d/dx [sin (x * ln x)] = cos (x * ln x) * (derivative of x * ln x).  We have to use the product rule to find the derivative of x * ln x:

 

d/dx (x * ln x) = x * (derivative of ln x) + (derivative of x) * ln x

= x (1/x) + 1(ln x)

= 1 + ln x.

 

Substituting this into our previous equation yields

 

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

= x(ln x + 1)cos(x*ln x) + sin(x*ln x)