NCERT Solutions for Class 12 Maths: Continuity and Differentiability Ex 5.4

Fundamental Derivatives: Exponential and Logarithmic Functions

Derivative of Exponential Function (e^x)

The function f(x) = e^x has a unique and remarkable property: its derivative is the function itself. d/dx (e^x) = e^x

Derivative of Natural Logarithmic Function (log x)

The derivative of the natural logarithm, log_e(x) or ln(x), where x > 0, is the reciprocal of x. d/dx (log x) = 1/x

Derivative of General Exponential Function (a^x)

For any positive base 'a' (where a ≠ 1), the derivative of a^x is the function itself multiplied by the natural log of the base. d/dx (a^x) = a^x * log(a)

Combining Rules: The Chain Rule with Exponential and Log Functions

Exercise 5.4 is all about applying the differentiation rules you already know to these new functions. The most important rule you'll use is the Chain Rule. You won't often be asked to differentiate just e^x or log x. Instead, you'll see composite functions like e^(sin x) or log(x² + 1).

Remember the chain rule: If y = f(g(x)), then dy/dx = f'(g(x)) * g'(x). Let's see how this works:

1. For Exponential Functions: If you have y = e^(g(x)), think of e^u as the outer function and u = g(x) as the inner function. The derivative is:

dy/dx = (derivative of e^u) * (derivative of g(x))
dy/dx = e^u * g'(x)
Substituting back u = g(x), we get: d/dx [e^(g(x))] = e^(g(x)) * g'(x).

1. For Logarithmic Functions: If you have y = log(g(x)), think of log u as the outer function and u = g(x) as the inner function. The derivative is:

dy/dx = (derivative of log u) * (derivative of g(x))
dy/dx = (1/u) * g'(x)
Substituting back u = g(x), we get: d/dx [log(g(x))] = (1 / g(x)) * g'(x) = g'(x) / g(x).

Mastering this pattern is the key to solving every problem in this exercise.

Worked Examples for Exercise 5.4

• Question 1: Differentiate e^x / sin(x) with respect to x. Step 1: Identify the rule to apply. The function is in the form of u/v, where u = e^x and v = sin(x). We must use the Quotient Rule: d/dx (u/v) = (v u' - u v') / v². Step 2: Find the derivatives of u and v. u = e^x => u' = d/dx(e^x) = e^x v = sin(x) => v' = d/dx(sin(x)) = cos(x) Step 3: Substitute these into the Quotient Rule formula. dy/dx = (sin(x) e^x - e^x cos(x)) / (sin(x))² Step 4: Simplify the expression. Factor out e^x from the numerator. Final Answer: dy/dx = e^x(sin(x) - cos(x)) / sin²(x)
• Question 2: Differentiate e^(sin⁻¹(x)) with respect to x. Step 1: Identify the rule to apply. This is a composite function of the form e^(g(x)), where g(x) = sin⁻¹(x). We need to use the Chain Rule. Step 2: Apply the Chain Rule for exponential functions. The rule is d/dx [e^(g(x))] = e^(g(x)) g'(x). Step 3: Find the derivative of the inner function, g(x). g(x) = sin⁻¹(x) g'(x) = d/dx(sin⁻¹(x)) = 1 / √(1 - x²) Step 4: Combine the parts. dy/dx = e^(sin⁻¹(x)) (1 / √(1 - x²)) Final Answer: dy/dx = e^(sin⁻¹(x)) / √(1 - x²)
• Question 3: Differentiate log(cos(e^x)) with respect to x. Step 1: Identify the structure of the function. This is a nested composite function: log(u), where u = cos(v), and v = e^x. We need to apply the Chain Rule twice. Step 2: Differentiate the outermost function. The outermost function is log(u). Its derivative is (1/u) u'. Here, u = cos(e^x). So, the derivative starts as: 1 / cos(e^x) d/dx(cos(e^x)). Step 3: Differentiate the next inner function. Now we need to find d/dx(cos(e^x)). This is another chain rule problem. The derivative of cos(v) is -sin(v) v'. Here, v = e^x. So, d/dx(cos(e^x)) = -sin(e^x) d/dx(e^x). Step 4: Differentiate the innermost function and combine everything. d/dx(e^x) = e^x. Putting it all together: dy/dx = [1 / cos(e^x)] [-sin(e^x)] [e^x] Step 5: Simplify the final expression. dy/dx = -e^x (sin(e^x) / cos(e^x)) = -e^x tan(e^x) Final Answer: dy/dx = -e^x * tan(e^x)

Exam Traps and Common Mistakes

In your board exams, questions from this section are designed to test your precision. Here's what to watch out for:

• Forgetting the Chain Rule: The most common mistake is seeing e^(2x) and writing its derivative as e^(2x). This is incorrect! You must multiply by the derivative of the inner function (2x), which is 2. The correct derivative is 2e^(2x).
• Misapplying the Log Rule: Similarly, for log(x² + 1), students often write 1 / (x² + 1). You must multiply by the derivative of x² + 1, which is 2x. The correct derivative is 2x / (x² + 1).
• Quotient Rule Errors: The quotient rule (v u' - u v') / v² has a minus sign. It's easy to accidentally swap the terms to u v' - v u', which gives the wrong sign. Always double-check the formula.
• Nested Functions: For functions like cos(log(x)), remember to work from the outside in. Differentiate cos, then log, then x, multiplying each derivative as you go. Don't try to do it all at once.

Practice Questions with Solutions

• Q: Differentiate y = sin(tan⁻¹(e⁻ˣ)) with respect to x. A: Step 1: Identify the nested functions. Let y = sin(u), u = tan⁻¹(v), and v = e⁻ˣ. We will use the chain rule repeatedly. Step 2: Find the derivatives of each part. dy/du = cos(u), du/dv = 1/(1+v²), dv/dx = e⁻ˣ (-1) = -e⁻ˣ. Step 3: Apply the chain rule: dy/dx = (dy/du) (du/dv) (dv/dx). Step 4: Substitute the functions back. dy/dx = cos(tan⁻¹(e⁻ˣ)) (1 / (1 + (e⁻ˣ)²)) (-e⁻ˣ). Final answer: dy/dx = -e⁻ˣ cos(tan⁻¹(e⁻ˣ)) / (1 + e⁻²ˣ).
• Q: Differentiate y = eˣ + eˣ² + ... + eˣ⁵ with respect to x. A: Step 1: The derivative of a sum is the sum of the derivatives. We can differentiate each term separately. dy/dx = d/dx(eˣ) + d/dx(eˣ²) + d/dx(eˣ³) + d/dx(eˣ⁴) + d/dx(eˣ⁵). Step 2: Apply the chain rule for each term of the form e^(g(x)). The derivative is e^(g(x)) g'(x). Step 3: Calculate each derivative: d/dx(eˣ) = eˣ d/dx(eˣ²) = eˣ² 2x d/dx(eˣ³) = eˣ³ 3x² d/dx(eˣ⁴) = eˣ⁴ 4x³ d/dx(eˣ⁵) = eˣ⁵ * 5x⁴ Step 4: Add all the derivatives together. Final answer: dy/dx = eˣ + 2xeˣ² + 3x²eˣ³ + 4x³eˣ⁴ + 5x⁴eˣ⁵.
• Q: Differentiate y = √(e^√x), where x > 0. A: Step 1: Rewrite the function using exponents: y = (e^(x^(1/2)))^(1/2). This is a nested chain rule problem. Step 2: Apply the chain rule from the outside in. The outermost function is u^(1/2). Its derivative is (1/2)u^(-1/2) u'. dy/dx = (1/2) (e^√x)^(-1/2) d/dx(e^√x). Step 3: Differentiate the next inner function, e^√x. The derivative is e^√x d/dx(√x). dy/dx = (1/2) (e^√x)^(-1/2) e^√x (1/(2√x)). Step 4: Simplify the expression. dy/dx = (1 / (2√(e^√x))) e^√x (1/(2√x)). Since e^√x / √(e^√x) = √(e^√x), we have: dy/dx = (√(e^√x)) / (4√x). Final answer: dy/dx = e^√x / (4√x √(e^√x)). A simpler form is √(e^√x) / (4√x).
• Q: Differentiate y = cos(log x + e^x), where x > 0. A: Step 1: Identify the composite function. Let y = cos(u) where u = log x + e^x. We will use the chain rule. Step 2: According to the chain rule, dy/dx = -sin(u) du/dx. dy/dx = -sin(log x + e^x) d/dx(log x + e^x). Step 3: Find the derivative of the inner function, u = log x + e^x. d/dx(log x + e^x) = d/dx(log x) + d/dx(e^x) = 1/x + e^x. Step 4: Substitute this back into the expression from Step 2. Final answer: dy/dx = -sin(log x + e^x) * (1/x + e^x).

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