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Differentiate the following w.r.t. x:`(xcosx)^x+(xsinx)^(1/x)` |
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Answer» Let `y = (xcosx)^x+(xsinx)^(1/x)` Let `u = (xcosx)^x and v = (xsinx)^(1/x)` Then, `y = u+v` `dy/dx = (du)/dx+(dv)/dx` Now, `u = (xcosx)^x` Taking log both sides, `log u = x(logx+logcosx)` Differentiating both sides, `1/u (du)/dx = (logx+cosx)*1+x(1/x+1/cosx(-sinx))` `1/u (du)/dx = (logx+cosx)+1-xtanx` `(du)/dx = (xcosx)^x(log(xcosx)+1-xtanx)->(1)` Now, `v = (xsinx)^(1/x)` Taking log both sides, `log v = 1/x(logx+logsinx)` Differentiating both sides, `1/v (dv)/dx = 1/x^2(x(1/x+1/sinx(cosx))-(logx+logsinx)*1)` `1/v(dv)/dx = 1/x^2(1+cotx -log(xsinx))` `(dv)/dx = (xsinx)^(1/x)1/x^2(1+cotx -log(xsinx))` So, `dy/dx = (du)/dx+(dv)/dx` `dy/dx = (xcosx)^x(log(xcosx)+1-xtanx)+(xsinx)^(1/x)((1+cotx -log(xsinx))/x^2)` |
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