Find dy/dx if y=x^tanx +[ (x^2+1)/2]^1/2
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let u=x^tanx. logu= xlog(tanx)
1/u.du/dx= x.1/tanx.sec^x+ log(tanx)
let v=[( x^2+1)/2]^1/2
dv/dx= 1/2[(x^2+1)/2]^-1/2.x
now find Dy/dx by du/dx+dv/dx
1/u.du/dx= x.1/tanx.sec^x+ log(tanx)
let v=[( x^2+1)/2]^1/2
dv/dx= 1/2[(x^2+1)/2]^-1/2.x
now find Dy/dx by du/dx+dv/dx
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