Differentiate the function and find the slope of the tangent line at the given value of the independent variable.
-$$\mathrm { s } = 2 \mathrm { t } ^ { 4 } - 2 \mathrm { t } ^ { 3 } , \mathrm { t } = - 1$$
A) 14
B) $- 14$
C) $- 2$
D) 2

Question

Quizplus · Accepted Answer

The derivative of $ s = 2t^4 - 2t^3 $ is $ s' = 8t^3 - 6t^2 $. Substituting $ t = -1 $ gives $ s'(-1) = 8(-1)^3 - 6(-1)^2 = -8 - 6 = -14 $.

Differentiate the Function and Find the Slope of the Tangent s=2t4−2t3,t=−1\mathrm { s } = 2 \mathrm { t } ^ { 4 } - 2 \mathrm { t } ^ { 3 } , \mathrm { t } = - 1s=2t4−2t3,t=−1

Differentiate the Function and Find the Slope of the Tangent $\mathrm { s } = 2 \mathrm { t } ^ { 4 } - 2 \mathrm { t } ^ { 3 } , \mathrm { t } = - 1$