## Calculus 10th Edition

The equation of the tangent line at the given point is $y=4x+10$
$f(x)=(x+2)(x^{2}+5)$ $,$ $(-1,6)$ First, use the product rule to evaluate the derivative of the function: $f'(x)=(x+2)(x^{2}+5)'+(x^{2}+5)(x+2)'=...$ Evaluate the indicated derivatives and simplify: $...=(x+2)(2x)+(x^{2}+5)(1)=2x^{2}+4x+x^{2}+5=...$ $...=3x^{2}+4x+5$ Substitute $x$ by $-1$ in the derivative found to obtain the slope of the tangent line: $m=3(-1)^{2}+4(-1)+5=3(1)-4+5=3-4+5=4$ Both the slope of the tangent line and a point through which it passes are now known. Use the point-slope formula of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point: $y-6=4[x-(-1)]$ $y-6=4(x+1)$ $y-6=4x+4$ $y=4x+4+6$ $y=4x+10$