JAMIA 2026 — Mathematics PYQ

Step 1: Simplify and Differentiate $y$

Given:

$$y={\cos }^{-1}x$$

Differentiating $y$ directly with respect to $x$:

$$\frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}$$

Step 2: Simplify and Differentiate $z$

Given:

$$z={\sin }^{-1}\sqrt{1-x^2}$$

Substitute $x=\cos \theta$ into the expression:

$$\sqrt{1-x^2}=\sqrt{1-{\cos }^2\theta }=\sqrt{{\sin }^2\theta }=\sin \theta$$

Now substitute this back into the function $z$:

$$z={\sin }^{-1}(\sin \theta )$$

$$z=\theta$$

Since $\theta ={\cos }^{-1}x$, we get:

$$z={\cos }^{-1}x$$

Now, differentiate $z$ with respect to $x$:

$$\frac{dz}{dx}=-\frac{1}{\sqrt{1-x^2}}$$

Step 3: Calculate $\frac{dy}{dz}$

Using our parametric formula:

$$\frac{dy}{dz}=\frac{\frac{dy}{dx}}{\frac{dz}{dx}}$$

Substitute the derivatives from Step 1 and Step 2:

$$\frac{dy}{dz}=\frac{-\frac{1}{\sqrt{1-x^2}}}{-\frac{1}{\sqrt{1-x^2}}}$$

Both the numerator and denominator are identical, so they cancel each other out completely:

$$\frac{dy}{dz}=1$$

Correct Answer:

(d) 1