Derivatives of Inverse Trigonometric and Parametric Functions

1. Differentiating Inverse Trigonometric Functions

You've already studied the properties of inverse trigonometric functions, but how do we find their rate of change? The answer lies in the implicit differentiation technique you learned in the previous lesson.

Let's take $y={\sin }^{-1}(x)$. This is equivalent to writing $x=\sin (y)$. If we differentiate this implicit relationship with respect to $x$, we get:

Solving for $\frac{dy}{dx}$ gives us:

Using the identity ${\cos }^2(y)+{\sin }^2(y)=1$, we can express $\cos (y)$ as $\sqrt{1-{\sin }^2(y)}$. Since $\sin (y)=x$, this becomes $\sqrt{1-x^2}$. Thus, we arrive at the famous result:

All six inverse trigonometric derivatives can be found this way. The following resource walks through each derivation.

After working through the derivations, the key is to commit the final formulas to memory. Here are the standard results, now combined with the chain rule for a general function $u(x)$:

• $\frac{d}{dx}({\sin }^{-1}u)=\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$
• $\frac{d}{dx}({\cos }^{-1}u)=-\frac{1}{\sqrt{1-u^2}}\cdot \frac{du}{dx}$
• $\frac{d}{dx}({\tan }^{-1}u)=\frac{1}{1+u^2}\cdot \frac{du}{dx}$
• $\frac{d}{dx}({\cot }^{-1}u)=-\frac{1}{1+u^2}\cdot \frac{du}{dx}$
• $\frac{d}{dx}({\sec }^{-1}u)=\frac{1}{|u|\sqrt{u^2-1}}\cdot \frac{du}{dx}$
• $\frac{d}{dx}({\csc }^{-1}u)=-\frac{1}{|u|\sqrt{u^2-1}}\cdot \frac{du}{dx}$

Notice the patterns: the derivatives of cos, cot, and csc functions are just the negative of their sin, tan, and sec counterparts.

The following video from The Organic Chemistry Tutor demonstrates how to apply these formulas with the chain rule in practice.

Simplification Before Differentiation

A crucial strategy for MHT-CET is to simplify complex inverse trigonometric expressions before you differentiate. Applying the chain rule directly to a complicated function like $y={\tan }^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ would be a nightmare. Instead, we can use a trigonometric substitution to simplify the argument.

The key is to recognize patterns that match trigonometric identities.

For the example above, the expression $\sqrt{1+x^2}$ suggests the substitution $x=\tan \theta$, which uses the identity $1+{\tan }^2\theta ={\sec }^2\theta$. Let's see how this plays out in a solved example.

2. Differentiating Parametric Functions

Next, we consider functions defined in parametric form. Instead of having $y$ as a direct function of $x$, both $x$ and $y$ are given as functions of a third variable, called a parameter (commonly $t$ or $\theta$). For example, the path of a projectile in physics can be described by its horizontal position $x(t)$ and vertical position $y(t)$ as functions of time.

Our goal is still to find $\frac{dy}{dx}$, the rate of change of $y$ with respect to $x$. We can do this using the chain rule:

Rearranging this to solve for $\frac{dy}{dx}$ gives us the core formula for parametric differentiation:

The following video provides an excellent, MHT-CET-focused explanation of this concept.

Differentiating One Function with Respect to Another

A very common MHT-CET question type asks you to "differentiate function $f(x)$ with respect to function $g(x)$". This is just a disguised parametric problem where $x$ itself is the parameter.

Let $u=f(x)$ and $v=g(x)$. The problem is asking for $\frac{du}{dv}$. Using the parametric rule with $x$ as the parameter, we get:

The same video shows exactly how to tackle this.

Conclusion

In this lesson, you've added two powerful techniques to your calculus toolkit, both of which are direct applications of the chain rule. You've seen how to differentiate inverse trigonometric functions and how to tackle parametric equations, including the common MHT-CET variant of differentiating one function with respect to another.

Key Takeaways:

• Derivatives of Inverse Trig Functions: These are a set of standard formulas. In problems, they are almost always used with the chain rule: $\frac{d}{dx}{f}^{-1}(u)=(\text{formula for }{f}^{-1})\cdot u^{\prime }$.
• Simplify First: For complex inverse trigonometric expressions, always look for an opportunity to use trigonometric identities or substitutions to simplify the expression before differentiating.
• Parametric Differentiation: For $x=f(t)$ and $y=g(t)$, the derivative is found by taking the ratio of individual derivatives with respect to the parameter: $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$.
• Function vs. Function: Differentiating $u(x)$ with respect to $v(x)$ is a parametric problem where $x$ is the parameter. The derivative is $\frac{du/dx}{dv/dx}$.

In our next lesson, we will explore logarithmic differentiation. This is a special technique used for functions that are otherwise very difficult or impossible to differentiate with our current rules, particularly those of the form $y=[f(x){]}^{g(x)}$.