I'm stuck trying to analytically prove that a partial derivative of a specific, lower defined function $C$ is negative. The context of this problem is actually a Black-Scholes market situation, where a price of a call option decreases as its strike increases.
For given positive constants $S, K, r, \sigma$ and $T$, we have: $$C(S,K,r, \sigma,T)=S \Phi(d_1)-Ke^{-rT}\Phi(d_2),$$ where $$d_1=\frac{\ln \frac{S}{K}+(r+\frac{1}{2}\sigma^2)T}{\sigma \sqrt{T}},$$ $$d_2=d_1-\sigma \sqrt{T}.$$
I have to prove that the function $C$ is decreasing if $K$ is increasing. First, I calculate the partial derivation: \begin{align} \frac{\partial C}{\partial K}&=S\frac{d \Phi(d_1)}{d (d_1)} \frac{\partial d_1}{\partial K}-e^{-rT}\Phi(d_2)-Ke^{-rT}\frac{d \Phi(d_2)}{d (d_2)}\frac{\partial d_2}{\partial K}\\ & = S \varphi(d_1)\frac{K}{\sigma \sqrt{T}}-e^{-rT}\Phi(d_2)-Ke^{-rT}\varphi(d_2)\frac{K}{\sigma \sqrt{T}}\\ & = K^2 \left( -e^{-rT} \frac{\varphi(d_2)}{\sigma \sqrt{T}} \right) + K \left( S \frac{\varphi(d_1)}{\sigma \sqrt{T}}\right) -e^{-rT}\Phi(d_2). \end{align} where $\Phi(x)$ is the standard normal cumulative, and $\varphi(x)$ standard normal density function.
This is more or less what I've got. I understand that I should somehow prove that the last expression is always non-negative, so I've tried calculating the determinant of the quadratic function of $K$, and I got $$D= \frac{S^2 (\varphi(d_1)^2)}{\sigma^2 T}-4 \frac{e^{-2rT }\varphi(d_2)\Phi(d_2)}{\sigma \sqrt{T}}.$$
And now I should prove that it is positive. No idea how? There is also a chance I misunderstood something and am leaving out some necessary conditions, I'm not sure.
Thanks for an insights on this, I really appreciate it.