The document Derivative of Antiderivative (Leibniz Rule) JEE Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.

All you need of JEE at this link: JEE

**Derivative of Antiderivative (Leibnitz Rule) **

If h(x) & g(x) are differentiable function of x then,

**Ex.15 Find the derivative of the function g(x) = **

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

**Ex.16 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.17 Find **

**Sol.** Let u = x^{4}. Then

**Ex.18 FInd the derivative of F(x) = **

**Sol.**

= (cos u) (3x^{2}) = (cos x^{3}) (3x^{2})

F'(x) = (cos x^{3}) (3x^{2}).**Ex.19 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) (x^{2 }+ x + 1)^{2} – (a + 1) (x^{2} + x + 1) (x^{2} – x + 1).

Now, f'(x) = 0 ⇒ (a – 1) (x^{2} + x + 1) – (a + 1) (x^{2} – x + 1) = 0 ⇒ x^{2} – ax + 1 = 0.

For distinct real roots D > 0 i.e. a^{2} – 4 > 0 ⇒ a^{2} > 4 ⇒ a ∈ (-∝, -2) U (2, ∝)

**Ex.20 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.21 Evaluate**** **

**Sol.**

**Ex.22 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 + x^{2}) > 0, the given improper integral can be interpreated as the area of the infinite region that lies under the curve y = 1/(1 + x^{2}) and above the x-axis (see Figure).

**Ex.23 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.24 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.25 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

Offer running on EduRev: __Apply code STAYHOME200__ to get INR 200 off on our premium plan EduRev Infinity!