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
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)
2 answers
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)
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)