Animated Solution for Mathematics - Differentiation: Let $f(x)=x^3+x^2{f}^{\prime }(1)+x{f}^{\prime \prime }(2)+f^{\prime \prime \prime }(3),x\in R$. Then $f^{\prime }(10)$ is equal to

My dear student, welcome to a classic JEE Advanced challenge. When you first look at the function $f(x)=x^3+x^2{f}^{\prime }(1)+x{f}^{\prime \prime }(2)+f^{\prime \prime \prime }(3)$, it is natural to feel a surge of anxiety. It looks like a self-referential loop, doesn't it?

But here is the secret: JEE problems often use notation to test your conceptual clarity, not just your ability to calculate. The terms $f^{\prime }(1)$, $f^{\prime \prime }(2)$, and $f^{\prime \prime \prime }(3)$ are not variables. They are the values of the derivatives at specific points; they are, quite simply, constants.

Let us peel back the layers. Since these terms are constants, let us assign them simple names: $f^{\prime }(1)=a$ $f^{\prime \prime }(2)=b$ $f^{\prime \prime \prime }(3)=c$

Now, our function becomes a friendly, standard cubic polynomial:

This is the moment the problem shifts from a terrifying functional equation to a manageable algebraic one.

Now, we need to find the values of $a$, $b$, and $c$. We do this by differentiating our new polynomial.

First, the first derivative:

Next, the second derivative:

Finally, the third derivative:

Notice how the variable $x$ vanishes in the third derivative? This is the beauty of polynomials.

Now, we use our definitions to solve for the constants. Since $f^{\prime \prime \prime }(x)=6$ for all $x$, it follows that:

Next, we use the second derivative:

Now, for the first derivative:

We now have a system of two linear equations: $b=12+2a$ and $a+b+3=0$. Substituting the first into the second:

Plugging $a=-5$ back into our equation for $b$:

We have our constants: $a=-5$, $b=2$, and $c=6$. Our derivative function is:

The question asks for $f^{\prime }(10)$. Substituting $x=10$:

The final answer is 202. You see? By breaking the problem down and trusting the process, we turned a daunting expression into a simple, elegant solution. Keep this mindset, and you will conquer any problem JEE throws at you.