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d/dx of e^x + 10^x + x^x . RD sharma shows the resultant answer as “ e^x + 10^x ln10 + x^x log(ex) .. How? as d/dx of x^x is x^x(1+lnx)

d/dx of e^x + 10^x + x^x . 
RD sharma shows  the resultant answer as 
“ e^x + 10^x ln10 + x^x log(ex) .. How? as d/dx of x^x is x^x(1+lnx) 

Grade:12

2 Answers

ng29
1698 Points
8 years ago
its easy look , u had problem in derivative of  xx , i will solve stepwise
let  xx = y  and we have to find  dy/dx ??
now take log on both sids and we get  lny = x lnx
now differentiate bot side wrt x 
(1/y)(dy/dx) = 1+lnx
dy/dx = y(1+lnx)
and we knoew that  y =xx  so  finally    dy/dx=xx(1+lnx)
 
 
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Satviki Pathak
40 Points
8 years ago
d/dx of x^x is x^x(1+lnx)  But  1 can be written as lne.
Therefore d/dx of x^x becomes x^x(lne+lnx).
Now, we use this logarithm property : logm +logn =logmn ( This property can be used only when base of both logm and logn is same) 
In our case, the base is same (that is e)
Therefore, x^x(lne+lnx) can be written as x^x ln(ex) 
 
 

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