Question

y = x^sinx + sinx^x find dy/dx

Answer 1

Answer:

Step-by-step explanation: y = x^sinx + sinx^x find dy/dx

let u = x^sinx and v = sinx^x  therefore, y = u + v

differentiating with respect to x, we get

dy/dx = du/dx + dv/dx             -----------(1)

Here, u = x^sinx

taking log both side

㏒u = ㏒x^sinx

logu = sinx logx

again differentiating with respect to x, we get

1/u du/dx = sinx (d/dx logx) + logx ( d/dx sinx )

1/u du/dx = sinx . 1/x + logx cosx

1/u du/dx = [sinx/x + log cosx ]

du/dx = u [sinx/x + logx cosx ]

du/dx = x^sinx [ sinx/x + logx cosx ]        -------------(2)

similarly,

v = sinx^x

differentiating with respect to x,  we get

1/v dv/dx = x cotx + logsinx

dv/dx = v [x cotx + logsinx ]

dv/dx = sinx^x [x cotx + logsinx ]       ----------------(3)

putting the value of du/dx from (2) and dv/dx from (3) in equation (1), we get

dy/dx = x^sinx [sinx/x + logx cosx ] + sinx^x [x cotx + logsinx ]