Derivative of Antiderivative (Leibniz Rule) | Mathematics (Maths) Class 12 - JEE PDF Download

Rule:

Solved Examples

Sol. Since f(t) =    is continuous, therefore g'(x) =

Ex.2 If F(t) = dx, find F'(1), F'(2), and F'(x).

Sol. The integrand in this example is the continuous function f defined by f(x) =

Ex.3 Find 

Sol. Let u = x⁴. Then

Ex.4 FInd the derivative of F(x) = 

Sol.

 = (cos u) (3x²) = (cos x³) (3x²)

F'(x) = (cos x³) (3x²).

Ex.5 Let f(x) = . Find the value of 'a' for which f'(x) = 0 has two distinct real roots.

Sol. Differentiating the given equation, we get f'(x) = (a – 1) (x² + x + 1)² – (a + 1) (x² + x + 1) (x² – x + 1).
Now, f'(x) = 0 ⇒ (a – 1) (x² + x + 1) – (a + 1) (x² – x + 1) = 0 ⇒ x² – ax + 1 = 0.
For distinct real roots D > 0 i.e. a² – 4 > 0 ⇒  a² > 4  ⇒  a ∈ (-∝, -2) U (2, ∝)

Ex.6 Show that for a differentiable function f(x), 

(where [ * ] denotes the greaetest integer function and n ε N)

Sol.

= – f(1) – f(2) – ........ – f(n – 1) – f(n)

Ex.7 Evaluate 

Sol.

Ex.8 Evaluate 

Sol.

We must now evaluate the integrals on the right side separately :

Since both of these integrals are convergent, the given integral is convergent and   Since 1/(1 + x²) > 0, the given improper integral can be interpreated as the area of the infinite region that lies under the curve y = 1/(1 + x²) and above the x-axis (see Figure). 

Ex.9 Find 

Sol.

We note first that the given integral is improper because f(x) = 1/√(x-2) has the vertical asymptote x = 2. Since the infinite discontinuity occurs at the left end point of [2, 5] 

Thus, the given improper integrat is convergent and, since the integrand is positive, we can interpret the value of the integral as the area of the shaded region in Figure.

Ex.10 Evaluate  

Sol. We know that the function f(x) = ln x has a vertical asymptote at 0 since  Thus, the given integral is improper and we have  

Now we integrate by parts with u = ln x, dv = dx, du = dx/x, and v = x

 =1 ln – t ln t – (1 – t)  = – t ln t – 1 + t

To find the limit of the first term we use I'Hopital's Rule :

Therefore     = –0 – 1 + 0 = –1

Figure shows the geometric interpretation of this result. The area of the shaded region above y = ln x and below the x-axis is 1.

Ex.11 Evaluate     (where [ * ] denotes the greatest integer function)

Sol.

for x > ln 2  ⇒  e^x > 2 ⇒ e^-x < 1/2 ⇒ 2e^–x < 1  ∴ 0 ≤ 2e^-x < 1 [2e^-x] = 0