Sigma Percentile
JEE Main 2023 (06 April Shift 1)
LEVELJEE Main

Animated Solution for Mathematics - Differentiation: If , then at is equal to:

Select Answer:

Visualized Solution

The Implicit Curve

  • Given Equation:
  • Target: Find at the point
  • This represents the slope of the tangent to the curve at .

Derivative of

  • Standard formula for variable base and exponent:
  • We will apply this to both and terms.

Differentiating the First Term

  • Differentiating with respect to :

Differentiating the Second Term

  • Differentiating with respect to :

Combine and Set to Zero

  • Add the derivatives and equate to the derivative of (which is ):

Substitute and

  • Substitute immediately to avoid messy algebra:

Expand the Equation

  • Expand the brackets carefully:

Group Terms

  • Group terms with on the left, move constants to the right:

Solve for

  • Isolate by dividing:
  • Factor out from numerator and denominator:

Final Log Transformation

  • Apply the power rule of logarithms:
  • Final Result:

The Sigma Insight: Techniques of Differentiation

Solution Diagram

Analyzing the Setup

Imagine you are standing before a complex, winding path on a graph defined by the equation . This is an implicit curve where and are locked in a dance, appearing in both the base and the exponent.
Our mission is to find the slope of the tangent line to this curve at the specific point . In the language of calculus, we are hunting for the value of at that coordinate.

The Power of the Formula

When you see a variable raised to a variable power, like , the standard power rule fails because the exponent is not a constant. We need a more robust tool.
We use the derivative formula for , where and are both functions of :
This formula is our secret weapon. It allows us to differentiate terms like and without the headache of taking logs of the entire equation.

Differentiating Term by Term

Let us apply this to our first term, . Keeping the constant aside, we identify and . The derivative becomes:
Now, we turn our attention to the second term, . Here, the base is and the exponent is . Applying the same logic, we get:
Adding these together and setting the result to the derivative of the constant (which is ), we obtain the full differentiated equation:

The Efficiency Shortcut

Instead of expanding this massive expression, we substitute and immediately. This collapses the variables into numbers.
Substituting these values, we get:
This simplifies beautifully to:
Expanding this, we get .

The Final Stretch

We group the terms containing on one side:
Solving for , we get:
Factoring out a from the numerator and denominator, we find:
Finally, applying the power rule of logarithms, , we transform into and into . Our final result is:

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