Some Important Results of Derivatives:

$\frac{d}{dx} (x) = 1$
$\frac{d}{dx} (constant) = 0$
$\frac{d}{dx} (x^n) = n{x}^{n-1}$
$\frac{d}{dx} (\log x) = \frac{1}{x}$
$\frac{d}{dx} (e^x) = e^x$
$\frac{d}{dx} (a^x) = {a^x} \log a$
$\frac{d}{dx} (\sqrt{x}) = \frac{1}{2\sqrt{x}}$

$\frac{d}{dx} (\sin x) = \cos x$
$\frac{d}{dx} (\cos x) = – \sin x$
$\frac{d}{dx} (\tan x) = \sec^{2} x$
$\frac{d}{dx} (\sec x) = \sec x . \tan x$
$\frac{d}{dx} (\cosec x) = – \cosec x . \cot x$
$\frac{d}{dx} (\cot x) = – \cosec^2 x$

$\frac{d}{dx} (sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}$
$\frac{d}{dx} (cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx} (tan^{-1}x) = \frac{1}{1+x^2}$
$\frac{d}{dx} (cot^{-1}x) = -\frac{1}{1+x^2}$
$\frac{d}{dx} (sec^{-1}x) = \frac{1}{x{\sqrt{x^2-1}}}$
$\frac{d}{dx} (cosec^{-1}x) = -\frac{1}{x{\sqrt{x^2-1}}}$

PRODUCT RULE:
$(I \times II)’ = I (II)’ + II (I)’$

QUOTIENT RULE:
$$\left( \frac{N}{D} \right )^’ = \frac{D(N)’ – N(D)’}{D^2}$$

LOGARITHM RULES:
$\log (m \times n) = \log m + \log n$
$\log \left ( \frac{m}{n} \right ) = \log m – \log n$
$\log (m)^n = n \log m$
$\log_a (b) = \frac{\log_x (b)}{\log_x (a)}$
$\log_e(e^x) = x$
$\log_a(a^x) = x$
$(e)^{\log_e{x}} = x$
$(a)^{\log_a{x}} = x$ Play Video

In this lecture, I am discussing remaining questions from NCERT Exercise 5.5 which are based on logarithmic differentiation of implicit functions.

Questions discussed in this lecture:

NCERT EXERCISE 5.5 (Logarithmic Differentiation)

Find $\frac{dy}{dx}$ of the functions given in Exercises 12 to 15.

Question 12. $x^y + y^x =1$

Question 13. $y^x = x^y$

Question 14. $(\cos x)^y = (\cos y)^x$

Question 15. $xy = e^{(x-y)}$

Question 16. Find the derivative of the function given by $f(x) = (1+x)(1+x^2)(1+x^4)(1+x^8)$ and hence find f′(1).

Question 17. Differentiate $(x^2-5x+8)(x^3+7x+9)$ in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?

Question 18. If u, v and w are functions of x, then show that $\frac{d}{dx}(u.v.w) = \frac{du}{dx}.v.w + u.\frac{dv}{dx}.w + u.v.\frac{dw}{dx}$ in two ways – first by repeated application of product rule, second by logarithmic differentiation.

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गिरे और गिरकर संभलते रहे,
हवाओं ने बहुत कोशिश की,
मगर चिराग आँधियों में भी जलते रहे।