Answer
$v^{\prime}(x)= (2.2x^{1.2}+1.2) \sec^2 (x^{2.2}+1.2x-1) $
Work Step by Step
We have: $v(x)= \tan (x^{2.2}+1.2x-1)$
We differentiate both sides with respect to $x$.
$v^{\prime}(x)=\dfrac{d}{dx} [ \tan (x^{2.2}+1.2x-1)] \\= \sec^2 (x^{2.2}+1.2x-1) \times (2.2x^{2.2-1}+1.2) $
Simplify to obtain:
$v^{\prime}(x)= (2.2x^{1.2}+1.2) \sec^2 (x^{2.2}+1.2x-1) $
