Here are the some questions for your practice on Leibnitz's Rule:
Q 1. Let $$ f(x) = \int_0^x (x^2-t^2)g(t)dt, $$ where g is a real valued continuous function on \( \mathbb {R}, f'(x) \) is equal to
\( (a) 0 \qquad (b) x^3 g(x) \)
\( (c) \int_0^x g(t)dt \qquad (d) 2x \int_0^x g(t)dt \)
Solution. Now we have to find f'(x). By Leibnitz's Rule
$$ \frac{d}{dx} \int_a^x F(x,t)g(t)dt = \int_a^x \frac{d}{dx} F(x,t)g(t)dt + F(x,x) \frac{dx}{dx} - F(x,a) \frac {da}{dx} $$
Here, \( F(x,t) = x^2-t^2 \) In the RHS of the formula we first \( F(x,t) \) differentiate w.r.t. 'x' then we will put upper limit and lower limit in place of 't' i.e., \( F(x,t)=F(x,x)\ and\ F(x,a) \) and further, differentiate upper and lower limit w.r.t. 'x' as given below:
$$ f'(x) = \int_0^x \frac {d}{dx} (x^2-t^2)g(t)dt + (x^2-x^2) \frac{dx}{dx} - (x^2-0^2) \frac {d0}{dx} $$
$$ f'(x) = \int_0^x 2x g(t)dt + 0 + 0 $$ $$ f'(x) = 2x \int_0^x g(t)dt $$
So, option d is the answer.
This is the way to find f'(x). Now you will solve the following questions given below:
Q2. The initial value problem corresponding to integral equation $$ y(x) = 1 + \int_0^x y(t)dt $$ is
$$ (a) y'-y = 0, y(0) = 1 \qquad (b) y' + y = 0, y(0) = 0 $$
$$ (c) y'-y = 0, y(0) = 0 \qquad (d) y' + y = 0, y(0) = 1 $$
Q3. The Integral equation $$ y(x) = \int_0^x (x-t) y(t)dt - x \int_0^1 (1-t)y(t)dt $$ is
$$ (a) y^"-y = 0, y(0) = y(1 ) = 0 \qquad (b) y^"-y = 0, y(0) = 0, y'(0) = 0 $$
$$ (c) y^"+y = 0, y(0) = 0, y(1) = 0 \qquad (d) y^"+y = 0, y(0) = 0 = y'(0) $$
Q4. The initial value problem $$ \frac{d^2y}{dx} + y = 0, y(0) = 1, y'(0) = 0 $$ is equivalent to Volterra integral equation
\( (a) y(x) = 1 + \int_0^x (t-x)y(t)dt \quad (b) y(x) = 1 + \int_0^x (t+x)y(t)dt \)
\( (c) y(x) = 1 + \int_0^x xty(t)dt \qquad (d) y(x) = 1 + \int_0^x (x-t)y(t)dt \)
{Hint: In this question, you can solve by discarding options by applying Liebnitz Rule.}
Q5. Let $$ f(x) = \frac{sin(x)}{1 + x^2}. $$ Let \( y^{\text{th}} \) denote the \( n^{\text{th}} \) derivative of \( f(x) at x = 0 \) then the value of \(y^{100} + 9900y^{98} \)
\( (a) 0 \qquad (b) -1 \)
\( (c) 100 \quad (d) 1729 \)
Q6. The limit $$\lim_{x \to 0}\frac{1}{sin^2(x)}\int_0^x sin^{-1}(t)dt $$ is
\( (a) \frac{1}{2} \qquad (b) \frac{1}{4} \)
\( (c) \frac{3}{2} \qquad (d) \frac{5}{2} \)