Find the tangent line of 10^x at(1;10)?

1 Answer
Aug 12, 2018

y=10ln10x+10(1-ln10)y=10ln10x+10(1ln10)
y approx 23.0236x-13.0236y23.0236x13.0236

Explanation:

f(x)=10^xf(x)=10x

lnf(x) = xln10lnf(x)=xln10

d/dx(lnf(x)) = ln10ddx(lnf(x))=ln10

1/f(x) * f'(x) = ln10

f'(x) = ln10*10^x

The slope of the tangent to f(x) at x=1 (m) is f'(1)

:. m= f'(1) = 10ln10

Let the tangent be: y=mx+c

Since (1,10) is a point on the tangent:

10 = 10ln10*1 + c

c= 10(1-ln10)

Hence our tangent is: y = 10ln10x+10(1-ln10)

y approx 23.0236x-13.0236