The Student Room Group
Well, 10^x = e^(x ln 10) = (e^x)^ln 10, and you know how to differentiate that.
Take logs

y=10xy = 10^{x}

ln(y)=ln(10x)ln(y) = \ln(10^{x})

ln(y)=xln(10)ln(y) = x \ln(10)

Then differentiate (using the product rule for the x ln10)

1ydydx=ln(10)\frac{1}{y}\frac{dy}{dx} = \ln(10)

dydx=yln(10)\frac{dy}{dx} = y \ln(10)

Then replace y back in

dydx=10xln(10)\frac{dy}{dx} = 10^{x} \ln(10)
Reply 3
what he said ^

ddxAx>(Ax)(lnA) \frac{d}{dx} A^x -> (A^x)(lnA)

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