#### Answer

$y=x+3$

#### Work Step by Step

The slope of the tangent at x=a
is the derivative of f(x), calculated in x=a.
The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).\\\\$
The point-slope equation is
$y-f(a)=f^{\prime}(a)(x-a)$
$f(x)=(x^{3}+7)^{2/3}$
To find $f^{\prime}(x)$, we need the chain rule:
$f^{\prime}(x)=\displaystyle \frac{2}{3}(x^{3}+7)^{-1/3}\cdot(x^{3}+7)^{\prime}$
$f^{\prime}(x)=\displaystyle \frac{2}{3}(x^{3}+7)^{-1/3}\cdot 3x^{2}$
$\displaystyle \color{blue}{f^{\prime}(x)=\frac{2x^{2}}{(x^{3}+7)^{1/3}}}$
$f(1)=(1+7)^{2/3}=8^{2/3}=4$
$f^{\prime}(1)=\displaystyle \frac{2}{(1+7)^{1/3}}=\frac{2}{2}=1$
Point-slope form for the line passing through $(1,4),$ with slope $1:$
$y-4=1(x-1)$
... solve for y (to transform to slope-intercept form)....
$y=x-1+4$
$y=x+3$