First of all, let's observe that the derivative of a sum is the sum of the derivatives. So,

#d/dx (x\ln(x) + 7x) = d/dx x\ln(x) + d/dx 7x#

The easy part is calculating that #d/dx 7x = 7 d/dx x#: for the power rule #d/dx x^n=nx^{n-1}#, applied with #n=1#, one has that #d/dx x = 1#, and so #d/dx 7x=7#.

As for #x\ln(x)#, we apply the product rule, which says that #(fg)'=f'g+fg'#. In your case, #f(x)=x# and #g(x)=\ln(x)#. We have that #f'(x)=1# and #g'(x)=1/x#. So, #d/dx x\ln(x) = 1\cdot \ln(x) + x \cdot 1/x#

Finally, the answer is

#d/dx (x\ln(x) + 7x) = \ln(x)+1+7=\ln(x)+8#