Answer
$y=7x-3$
Work Step by Step
For x=1, the point on the graph is $(1, f(1))$.
The slope of the tangent at x=1 is $m=f^{\prime}(1).$
The point-slope equation of the tangent line is
$y-y_{1}=m(x-x_{1})$
$y-f(1)=f^{\prime}(1)(x-1)$
$f(1)=( 1+1)(1+1)=4$
For $f^{\prime}(x)$, use the product rule,
$\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\prime}(x)g(x)+f(x)g^{\prime}(x)$.
$=(x^{0.5}+1)^{\prime}(x^{2}+x)+(x^{0.5}+1)(x^{2}+x)^{\prime}$
$f^{\prime}(x)=(0.5x)(x^{2}+x)+(x^{0.5}+1)(2x+1)$
... evaluate at x=1 ...
$f^{\prime}(1)=0.5(1+1)+(1+1)(2+1)$
$=1+6=7$
An equation for the tangent line is
$y-4=7(x-1)$
$y=7x-7+4$
$y=7x-3$
