Calculus

Could someone please explain this to me?

Differentiate.

h(u)= 10^(sqrt(u))

h'(u)= 10^(sqrt(u))*log(10)(d/du(sqrt(u)))

h'(u)= 10^(sqrt(u))*((1)/(2*sqrt(u)))*log(10)

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  1. let x = √u = u^(1/2) , let h(u) = y

    then y = 10^x
    take ln of both sides
    ln y = ln 10^x
    ln y = xln10
    (dy/dx) / y= ln10
    dy/dx = y ln10
    = (10^x)(ln10)

    back to x = u^(1/2)
    dx/du = (1/2)u^(-1/2)

    then (dy/dx) (dx/du) = [(10^x)(ln10)(1/2)(u^(-1/2)
    dy/du = (10^√u)(ln10)/(2√u)

    you have log10, it should be ln10

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  2. Note that the "log(10)" above refers to natural log of 10.

    There was a time before calculators when log10(x) was used to help do multiply/divide operations. To avoid confusion, natural log was denoted by ln(x). It has stayed since.

    However, log(x) continues to be used for natural log in higher mathematics.

    Start with
    y=10^x
    take log10 on both sides:
    log10y=x
    differentiate:
    (1/(y*ln(10)))(dy/dx)=1
    dy/dx=y*ln(10)
    dy/dx = 10^x ln(10)

    The rest is just the chain rule, for example,
    y=sin(x)^4
    dy/dx = 4sin(x)^3*d(sin(x))/dx
    =4sin(x)^3 * cos(x)

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