How do you differentiate #s(x)=log_3(x^2+5x)#?

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Answer 1 · 2016-06-14T23:55:19

#s^'(x)=(2x+5)/((x^2+5x)ln(3))#

We can rewrite #s(x)# using the change of base formula, which states that

#log_a(b)=log_c(b)/log_c(a)#

Here, we will choose a base of #e# because differentiation with the natural logarithm is simple.

#s(x)=ln(x^2+5x)/ln(3)#

When differentiating this, note that #s(x)# is really the function #ln(x^2+5x)# multiplied by the constant #1/ln(3)#.

To find the derivative of #ln(x^2+5x)#, we will use the chain rule.

Since #d/dx(ln(x))=1/x#, we see that #d/dx(ln(u))=1/u*u^'.#

Thus:

#s^'(x)=1/ln(3)*1/(x^2+5x)*d/dx(x^2+5x)#

#s^'(x)=(2x+5)/((x^2+5x)ln(3))#