Answer
\begin{align*}
(a)&\frac{\partial k}{\partial a} =5 b^{3}\\
(b)&\frac{\partial k}{\partial b} =15 a b^{2}+7\left(1.4^{b}\right)\ln (1.4)\\
(c)&\frac{\partial k}{\partial b}\bigg|_{a=6} = 90 b^{2}+7\left(1.4^{b}\right)\ln (1.4)\\
\end{align*}
Work Step by Step
Given $$k(a, b)=5 a b^{3}+7\left(1.4^{b}\right)$$
Since
\begin{align*}
(a)&\frac{\partial k}{\partial a} =5 b^{3}\\
(b)&\frac{\partial k}{\partial b} =15 a b^{2}+7\left(1.4^{b}\right)\ln (1.4)\\
(c)&\frac{\partial k}{\partial b}\bigg|_{a=6} = 90 b^{2}+7\left(1.4^{b}\right)\ln (1.4)\\
\end{align*}
