Xrfgtjtk

I want to differentiate this with respect to $eta$:

$$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}$$

Does this solution look correct? I tried going through with the quotient rule, but it is a bit messy, so I was hoping someone could check my math

$$C'(eta) = frac{left[3left(1-frac{eta}{eta_c}right)sumlimits_{k=1}^{3} B_kkleft(frac{eta}{eta_c}right)^{k-1}frac{1}{eta_c}right]-left[left(1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^kright)left(-frac{3}{eta_c}right)right]}{left[3left(1-frac{eta}{eta_c}right)right]^2}$$

You could have made life a bit simpler writing
$$C(eta) = frac{1+sumlimits_{k=1}^{3} B_kx^k}{3left(1-xright)}qquad text{where}qquad x=frac{eta}{eta_c}$$ and use
$$frac{dC(eta)}{deta} = frac{d C(eta)}{dx}frac{dx}{deta}=frac{1}{eta_c}frac{d C(eta)}{dx}$$ For sure, your result is correct but, since $k$ has only a few values, you could simplity the final result to get
$$C'(eta)=frac{1}{eta_c}frac{1+B_1+2 B_2 x+(3 B_3-B_2) x^2-2 B_3 x^3 }{3(1-x)^2}$$

Here's how you could differentiate this particular function:

$$
C'(eta) = left[frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3left(1-frac{eta}{eta_c}right)}right]'=
left[frac{1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}right]'=\
left[left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left[3left(1-frac{eta}{eta_c}right)right]^{-1}right]'=\
left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)'left[3left(1-frac{eta}{eta_c}right)right]^{-1}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)left(left[3left(1-frac{eta}{eta_c}right)right]^{-1}right)'=\
frac{B_{1}frac{1}{eta_c}+B_{2}frac{2eta}{eta_c^2}+B_{3}frac{3eta^2}{eta_c^3}}{3left(1-frac{eta}{eta_c}right)}+left(1+B_{1}frac{eta}{eta_c}+B_{2}frac{eta^2}{eta_c^2}+B_{3}frac{eta^3}{eta_c^3}right)(-1)left[3left(1-frac{eta}{eta_c}right)right]^{-2}left(3-3frac{eta}{eta_c}right)'=\
frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{left[3left(1-frac{eta}{eta_c}right)right]^2}cdotfrac{3}{eta_c}=\
frac{sumlimits_{k=1}^{3}B_kfrac{keta^{k-1}}{eta_c^k}}{3left(1-frac{eta}{eta_c}right)}+frac{1+sumlimits_{k=1}^{3} B_kleft(frac{eta}{eta_c}right)^k}{3eta_cleft(1-frac{eta}{eta_c}right)^2}.
$$

The answer I get appears to be equivalent to what you got.