Find the slope of the tangent line to the curve $y = \frac { 4 } { 3 \left( x ^ { 3 } - 2 x ^ { 2 } + 3 x - 5 \right) } \text { at } \left( 2 , \frac { 4 } { 3 } \right)$
A) - $\frac { 8 } { 3 }$
B) - $\frac { 28 } { 3 }$
C) $\frac { 8 } { 3 }$
D) $\frac { 4 } { 3 }$
E) $\frac { 2 } { 3 }$

Question

Quizplus · Accepted Answer

To find the slope of the tangent line to the curve at a given point, we need to differentiate the given function with respect to $x$ and then substitute the $x$-coordinate of the point into the derivative. The derivative of $y = \frac { 4 } { 3 \left( x ^ { 3 } - 2 x ^ { 2 } + 3 x - 5 \right) }$ using the chain rule is $y' = - \frac { 4(3x^2 - 4x + 3) } { 3(x^3 - 2x^2 + 3x - 5)^2 }$. Substituting $x = 2$ into this derivative gives $y'(2) = - \frac { 4(3(2)^2 - 4(2) + 3) } { 3(2^3 - 2(2)^2 + 3(2) - 5)^2 } = - \frac { 28 } { 3 }$.

Find the Slope of the Tangent Line to the Curve y=43(x3−2x2+3x−5) at (2,43)y = \frac { 4 } { 3 \left( x ^ { 3 } - 2 x ^ { 2 } + 3 x - 5 \right) } \text { at } \left( 2 , \frac { 4 } { 3 } \right)y=3(x3−2x2+3x−5)4​ at (2,34​)

Find the Slope of the Tangent Line to the Curve $y = \frac { 4 } { 3 \left( x ^ { 3 } - 2 x ^ { 2 } + 3 x - 5 \right) } \text { at } \left( 2 , \frac { 4 } { 3 } \right)$