 Hello and welcome to the session. In this session we discuss the following question which says differentiate the following function with respect to x. x to the power sin x plus sin x to the power cos x. Let's proceed with the solution now. We take let y be equal to the given function that is x to the power sin x plus sin x to the power cos x. Then we take let u be equal to x to the power sin x and v be equal to sin x to the power cos x. Then we have y would be equal to u plus v. This means dy by dx is equal to du by dx plus dv by dx. Let this be equation one. Now first let's consider u equal to x to the power sin x. Now taking log on both sides we get log u is equal to sin x into log x. Now differentiating both sides with respect to x that is we have d by dx of log u is equal to d by dx of sin x into log x. This further gives us 1 upon u du by dx is equal to sin x into d by dx of log x plus log x into d by dx of sin x. Further 1 by u into du by dx is equal to sin x into 1 upon x that is d by dx of log x is 1 upon x plus log x into d by dx of sin x is cos x. Further du by dx is equal to u into sin x upon x plus sin x into log x. Now putting the value for u as x to the power sin x we get du by dx is equal to x to the power sin x whole multiplied by sin x upon x plus cos x into log x. Now let this be equation 2. Next we consider v equal to sin x to the power cos x. Now taking log on both sides we get log v is equal to cos x into log sin x. Now differentiating both sides with respect to x we get d by dx of log v is equal to d by dx of cos x into log of sin x. Further we get 1 upon v into d v by dx is equal to cos x into d by dx of log sin x plus log sin x into d by dx of cos x. This means we have 1 upon v into d v by dx is equal to cos x into 1 upon sin x into d by dx of sin x that is cos x plus log of sin x into d by dx of cos x which is minus sin x. This gives us d v by dx is equal to v into now cos x upon sin x is cos x into cos x minus sin x into log of sin x. Now putting the value for v as sin x to the power cos x we get d v by dx is equal to sin x to the power cos x. This whole multiplied by cos x into cos x minus sin x into log sin x. Let this be equation 3. Now from equation 1 we have dy by dx is equal to du upon dx plus dv upon dx. So now next we have using equations 2 and 3 in equation 1 we get dy by dx is equal to du by dx which is x to the power sin x whole multiplied by sin x upon x plus cos x into log x plus sin x to the power cos x whole multiplied by cos x into cos x minus sin x into log of sin x. Since we have taken the given function to be y so we were supposed to find dy by dx which is this that we have got the required solution. So this completes the session hope you have understood the solution of this question.