Answer to Question #214517 in Calculus for sbuda

Question #214517

Differentiate the following with regard to x

y=ln( square toot X multiple e^x^2)


1
Expert's answer
2021-07-08T12:02:35-0400

y=ln(x×ex2)y=ln( \sqrt {x} \times e^{x^2})


Let u=x×ex2    y=ln(u)    dydu=1uu=\sqrt {x} \times e^{x^2}\implies y=ln(u)\implies {dy\over du}={1\over u}


Let f=x=x12    f=12x121=12x12f=\sqrt x=x^{1\over 2} \implies f'={1\over 2}x^{{1\over 2}-1}={1\over 2}x^{-{1\over 2}}


Let g=ex2    g=2xex2g=e^{x^2} \implies g'=2xe^{x^2}


We then use product rule of differentiation to differentiate u=x×ex2u=\sqrt {x} \times e^{x^2} with respect to xx


dudx=fg+fg{du \over dx}=f'g+fg'


=12x12ex2+x(2xex2)={1\over 2}x^{-{1\over 2}}e^{x^2}+\sqrt x(2xe^{x^2})



We then use chain rule of differentiation to differentiate y=ln(x×ex2)y=ln( \sqrt {x} \times e^{x^2}) with respect to xx


dydx=dydu×dudx{dy\over dx}={dy\over du}\times{du\over dx}


=1u×(12x12ex2+x(2xex2))={1\over u}\times({1\over 2}x^{-{1\over 2}}e^{x^2}+\sqrt x(2xe^{x^2}))


=1x×ex2×(12x12ex2+x(2xex2))={1\over \sqrt {x} \times e^{x^2}}\times({1\over 2}x^{-{1\over 2}}e^{x^2}+\sqrt x(2xe^{x^2}))


=1x×ex2×(12(x)ex2+x(2xex2))={1\over \sqrt {x} \times e^{x^2}}\times({1\over 2(\sqrt x)}e^{x^2}+\sqrt x(2xe^{x^2}))

=ex22xex2+x(2xex2x(ex2)={e^{x^2}\over 2xe^{x^2}}+{\sqrt x(2xe^{x^2}\over \sqrt x(e^{x^2})}

=12x+2x={1\over 2x}+2x


dydx=1+4x22x\therefore {dy\over dx}={1+4x^2\over 2x}


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