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प्रश्न
Differentiate xsinx+(sinx)cosx with respect to x.
उत्तर
`y=x^(sinx)+(sinx)^cosx`
`=>y=e^(sinxlogx)+e^(cosxlogsinx)`
Differentiating both sides with respect to x, we get
`dy/dx=e^(sinx logx)xx d/dx(sinxlogx)+e^(cosxlog sinx)xxd/dx(cosx log sinx)`
`dy/dx=x^(sinx)(sinxxx1/x+logx xx cosx)+(sin)^(cosx)(cosx xx 1/sinx xx cosx +log sinx)`
`dy/dx=x^(sinx)((sinx)/x+cosx logx)+(sinx)^(cosx)((cos^2x)/(sinx)-sinx log sinx)`
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