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

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Answer 1 · 2018-08-12T13:26:21

#y=10ln10x+10(1-ln10)#
#y approx 23.0236x-13.0236#

#f(x)=10^x#

#lnf(x) = xln10#

#d/dx(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#