Find F ' x if F x = x x 2 x 3 1 2 x 3 x 2 0 2 6 x

Given, F x = x x 2 x 3 1 2 x 3 x 2 0 2 6 x ⇒ F ' x = 1 2 x 3 x 2 1 2 x 3 x 2 0 2 6 x + x x 2 x 3 0 2 6 x 0 2 6 x + x x 2 x 3 1 2 x 3 x 2 0 0 6 The first two determinants have two rows same. Hence, their values will be 0 . Expanding the third determinant along third row, we get F ' x = 6 x 2 x - 1 x 2 ⇒ F ' x = 6 2 x 2 - x 2 = 6 x 2

Find y '' for the following functions y = x 2 sin 1 x ; x ≠ 0 0 ; x = 0 . Is there y '' 0 ?

Given: y = x 2 sin 1 x ; x ≠ 0 0 ; x = 0 For x ≠ 0 d y d x = 2 x sin 1 x - cos 1 x ⇒ d 2 y d x 2 = 2 sin 1 x - 2 x cos 1 x - 1 x 2 sin 1 x ∵ d d x f x g x = f ' x g x + f x g ' x d d x x n = n x n - 1 d d x f g x = g ' x f ' g x At x = 0 , y ' ' 0 does not exists.

f x = x n show that f 1 + f ' 1 1 ! + f 2 1 2 ! + … + f n 1 n ! = 2 n .

We have, f x = x n ⇒ f ' x = n x n - 1 ⇒ f ' ' x = n n - 1 x n - 2 ⇒ f ' ' ' x = n n - 1 n - 2 x n - 3 Similarly, f n x = n n - 1 n - 2 . . . . 3 · 2 · 1 = n ! So, f 1 + f ' 1 1 ! + f 2 1 2 ! + … + f n 1 n ! = 1 + n + n n - 1 2 ! + n n - 1 n - 2 3 ! + . . . + n ! n ! = 1 + 1 n = 2 n Since, 1 + x n = 1 + n x + n n - 1 2 ! x 2 + n n - 1 n - 2 3 ! x 3 + . . . + n n - 1 n - 2 . . . . 3 · 2 · 1 n ! x n

If y = x 2 e - x a . Show that f n 0 = - 1 n n n - 1 a n - 2 n ≥ 2 .

Given: y = x 2 e - x a ⇒ f ' x = 2 x - x 2 a e - x a ⇒ f ' ' x = 2 - 4 x a + x 2 a 2 e - x a ⇒ f ' ' ' x = - 6 a + 6 x a 2 - x 2 a 3 e - x a ⇒ f ' ' ' ' x = 12 a 2 - 8 x a 3 + x 2 a 4 e - x a ⇒ f ' ' ' ' ' x = - 20 a 3 + 10 x a 4 - x 2 a 5 e - x a Proceeding in this manner, we have f n x = - 1 n n n - 1 a n - 2 + - 1 n - 1 2 n x a n - 1 + - 1 n x 2 a n e - x a ; n ≥ 2 Now, hence, f n 0 = - 1 n n n - 1 a n - 2 .

E M B I B E

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 2 - Differentiation of Functions>Additional Problems>Q 2.7.11.

MEDIUM

Mathematics

IMPORTANT

Earn 100

Prove that the derivative of a differentiable even function is an odd function, and the derivative of an odd function is an even function. Give a geometric explanation to these facts.

Important Questions on Differentiation of Functions

EASY

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 2 - Differentiation of Functions>Additional Problems>Q 2.7.12.

Prove that the derivative of a periodic function with period

T

is a periodic function with period

T

.

MEDIUM

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 2 - Differentiation of Functions>Additional Problems>Q 2.7.13.

Find

F^{'} (x)

if

F (x) = |\begin{matrix} x & x^{2} & x^{3} \\ 1 & 2 x & 3 x^{2} \\ 0 & 2 & 6 x \end{matrix}|

MEDIUM

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 2 - Differentiation of Functions>Additional Problems>Q 2.7.14.

Find the derivative of the function

y = x |x|

. Sketch the graphs of the given function and its derivative.

MEDIUM

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 2 - Differentiation of Functions>Additional Problems>Q 2.7.15.

Suppose we have a composite function $y = f (u)$ , where $u = φ (x)$ . Among what points should we look for points at which the composite function may have no derivative?

Does the composite function always have no derivative at these points? Consider the function $y = u^{2}, u = |x|$ .

MEDIUM

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 2 - Differentiation of Functions>Additional Problems>Q 2.7.16.

Find