Answer:
sec² [㏑(ax + b)] X a/(ax +b)
Explanation:
chain rule is if y = f(c) and z = f(e), and y = f[f(e)], then:
dy/de = dy/dc X dc/de
This might look complicated. Much easier with example, as in this question.
y = tan [㏑(ax + b)]
y = tan (u),
where u =㏑(v), v = ax + b
so we have y = tan (u), u = ㏑(v), v = ax + b.
dy/du = sec² (u), du/dv = 1/v, dv/dx = a.
dy/dx = dy/du X du/dv X dv/dx
= sec² (u) X (1/v) X a
= sec² [㏑(v)] X (1/(ax + b)) X a
= sec² [㏑(ax + b)] X a/(ax + b)