Question #a787b

2 Answers
Oct 22, 2016

d/dx(log(xe^x)/log(5))=(x+1)/(xlog(5))

Explanation:

Another way to do this is to simplify at first using the rule log(AB)=log(A)+log(B):

log(xe^x)/log(5)=(log(x)+log(e^x))/log(5)=(log(x)+x)/log(5)

Don't forget that log(5) is a constant, so we can bring it out when differentiating:

d/dx((log(x)+x)/log(5))=1/log(5)d/dx(log(x)+x)

The derivative of log(x) is d/dxlog(x)=1/x. The derivative of x is 1.

1/log(5)d/dx(log(x)+x)=1/log(5)(1/x+1)=1/log(5)((1+x)/x)

Combining this all:

d/dx(log(xe^x)/log(5))=(x+1)/(xlog(5))

Oct 22, 2016

(1+x) / (xln5)

Explanation:

STEP 1: Rewrite the expression using logarithm rules
If we remember our logarithm rules, we will recall that log_bx = logx/logb.
Using the logarithm rule above, we can rewrite log(xe^x)/log5 as log_5(xe^x).

STEP 2: Differentiate
Recall our differentiation rule for logarithms: d/dx log_ax=1/(xlna)

Combining our log derivative rule with the chain rule, we get:
d/dx log_5(xe^x) = 1/((xe^x)ln5) * d/dx(xe^x)
To take the derivative of xe^x we must remember to apply the product rule:
1/((xe^x)ln5) * d/dx(xe^x) = 1/((xe^x)ln5) * [1*e^x + x*e^x]

STEP 3: Simplify
1/((xe^x)ln5) * [1*e^x + x*e^x]=[e^x + xe^x] / ((xe^x)ln5) = (1+x) / (xln5)