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STEP I 1994 question 2 solution

From The Student Room

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1994 question 2 solution

Part i) y = x^a

\frac{\text{d}y}{\text{d}x} = ax^{(a-1)}


Part ii) \ y = a^x}

\ln y = \ln(a^x)

\ln y = x \ln a

\frac{\text{d}y}{\text{d}x} \frac{1}{y} = \ln a

\frac{\text{d}y}{\text{d}x} = y \ln a

\frac{\text{d}y}{\text{d}x} = a^x \ln a


Part iii) y = x^x

\ln y = \ln (x^x)

\ln y = x \ln x

\frac{\text{d}y}{\text{d}x} \frac{1}{y} = \ln x + 1

\frac{\text{d}y}{\text{d}x} = x^x( \ln x + 1)


Part iv) y = x^{x^x}

\ln y = \ln (x^{x^x})

\ln y = x^x \ln x

\frac{\text{d}y}{\text{d}x} \frac{1}{y} = x^x (\ln x + 1) \lnx + \frac{x^x}{x} (From part iii)

\frac{\text{d}y}{\text{d}x} \frac{1}{y} = x^x ((\ln x)^2 + \ln x) + x^{(x-1)}

\frac{\text{d}y}{\text{d}x} = yx^x(( \ln x)^2 + \ln x + \frac{1}{x})

\frac{\text{d}y}{\text{d}x} = x^{(x^x + x)}(( \ln x)^2 + \ln x + \frac{1}{x})


Part v) y = (x^x)^x

y = x^{x^2}

\ln y = \ln (x^{x^2})

\ln y = x^2 \ln x

\frac{\text{d}y}{\text{d}x} \frac{1}{y} = 2x \ln x + x

\frac{\text{d}y}{\text{d}x} = x^{x^2} (2x \ln x + x)

\frac{\text{d}y}{\text{d}x} = x^{(x^2+1)} (2 \ln x + 1)

Solution by zrancis

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