Let g(x)=xln(x). We know g(e)=e. Use a derivative to estimate g(e+0.1). You may leave an unsimplified numerical answer. I understand how to plug g(e+0.1) in, but not using the derivative.
If \(\displaystyle h \approx 0 \) then \(\displaystyle g'(x_0)\approx \dfrac{g(x_0+h)-g(x_0)}{h} \) then rearranging we get
\(\displaystyle g(x_0+h)\approx h\cdot g'(x_0)+g(x_0)\).
\(\displaystyle g(x)=x\log(x)~ \Rightarrow~g'(x)=\log(x)+1\) Thus let \(\displaystyle x_0=e~\&~h=0.1\).
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