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The Derivative of y=x^sinx...
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did you try logarithmic differentiation
When you do this:\[ x^{\sin x} = (e^{\ln x})^{\sin x} = e^{\ln x \sin x} \]It becomes a bit easier.
is the derivative cosx/x
Not quite
is this correct (sinx)x^(sinx - 1)(cosx)
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stow your working ill check
show*
yes
\[logy=\log(x)^{sinx} \rightarrow logy=sinxlogx\]
diff.w.r.t.x
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yep now @denizen continue
hmm
(sin x )(cos x)(x^(sin x -1))=1/2 sin (2x) x^(sinx -1)
\[i/y dy/dx=sinx/x+xcosx\]
log y=log(x^sin x) log y=sin x .log x diff. with respect to x (1/y)(dy/dx)=cos x.log x + sin x /x) dy/dx=(x^sin x) (cos x .log x + sin x/x) it's the answers of prob.
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