Answer to Question #187815 in Differential Equations for Alice

Question #187815

Find dy/dx and simply the result, if possible.(with solution)

A.     y=√x -(1/√x)

B.     y=x^2+π^2+x^π

C.     y=(sin x-1)/(cos x)

D.     y=x^2 sec x

E.      y=(1)/(e^× +2)


1
Expert's answer
2021-05-07T11:31:02-0400

A)

Y=x+1xY=\sqrt{x}+\dfrac{1}{\sqrt{x}}

dydx=12x+12x3\dfrac{dy}{dx}=\dfrac{1}{2\sqrt x}+\dfrac{1}{2\sqrt[3]{x}}


B)

Y= x2+π2+xπY=\ x^{2}+\pi^{2}+x^{\pi}

dydx=2x+πxπ1\dfrac{dy}{dx}=2x+\pi{x}^{\pi-1}


C)

Y=sinxcosx1Y=\dfrac{sinx}{cosx-1}

dydx=cosx×cosx(sinx1)(sinx)(cosx1)2\dfrac{dy}{dx}=\dfrac{cosx\times cosx-(sinx-1)(-sinx)}{(cosx-1)^2}

dydx=cos2x+sin2xsinx(cosx1)2\dfrac{dy}{dx}=\dfrac{cos^{2}x+sin^{2}x-sinx}{{{(cosx-1}})^{2}}


dydx=1sinx(cosx1)2\dfrac{dy}{dx}=\dfrac{1-sinx}{(cosx-1)^{2}}


D)

Y=x2secxY=x^{2}secx

dydx=2×x×secx+secx×tanx×x2\dfrac{dy}{dx}=2\times x\times secx+secx\times tanx \times x^{2}


=xsecx(xtanx+2)

E)

Y=1ex+2Y=\dfrac{1}{e^{x}+2}

dydx=ex+2×0+1×ex(ex+2)2=ex(ex+2)2\dfrac{dy}{dx}=\dfrac{e^{x}+2\times0+1\times e^{x}}{(e^x+2)^2}=\dfrac{e^{x}}{(e^x+2)^2}




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