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}}}

(I \times II)’ = I (II)’ + II (I)’

\left( \frac{N}{D} \right )^’ = \frac{D(N)’ – N(D)’}{D^2}

\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

Continuity and Differentiability Lecture 8
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Continuity and Differentiability Lecture 8
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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.

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