Find $\frac{dy}{dx}$ for the equation $ax + by^2 = \cos y$.

Consider the non-constant differentiable function $f$ of one variable which obeys the relation $\frac{f(x)}{f(y)}=f(x-y)$. If $f^{\prime}(0)=p$ and $f^{\prime}(5)=q$, then $f^{\prime}(-5)$ is

If $y = \sqrt{\cos x^2 + \sqrt{\cos x^2 + \sqrt{\cos x^2 + \dots \infty}}}$ and $\frac{dy}{dx} = \frac{f(x)}{2y - 1}$, then $\int f(x) dx = \dots$

If $\cos y = x \cos (a+y)$ with $\cos a \neq \pm 1,$ prove that $\frac{dy}{dx} = \frac{\cos^2 (a+y)}{\sin a}$.

Let $f$ be a differentiable function such that $(x - y) f(x + y) - (x + y) f(x - y) = 4xy(x^2 - y^2)$ and $f(1) = 2$. Then the area enclosed by $\frac{|f(x) - x|^{1/3}}{17} + \frac{|f(y) - y|^{1/3}}{2} \le \frac{1}{4}$ is

If $\sqrt{y + x} + \sqrt{y - x} = c$ and $\frac{dy}{dx} = K - \sqrt{\frac{y^2}{x^2} - 1}$, then the value of $K$ is