In the previous section, we completed a discussion on Rolle's theorem and mean value theorem. In this section, we will see some miscellaneous examples.
Solved example 21.66
Differentiate w.r.t x the following functions:
(i) $\rm{\sqrt{3x+2}~+~\frac{1}{\sqrt{2x^2 + 4}}}$
(ii) $\rm{e^{\sec^2 x}~+~3 \cos^{-1} x}$
(iii) $\rm{\log_7 (\log x)}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\sqrt{3x+2}~+~\frac{1}{\sqrt{2x^2 + 4}}} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\left(3x+2 \right)^{1/2} ~+~\left(2x^2 + 4 \right)^{-1/2}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{1/2 \left(3x+2 \right)^{-1/2}.3 ~+~(-1/2)\left(2x^2 + 4 \right)^{-3/2}.4x} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{3}{2 \sqrt{3x+2}}~-~\frac{2x}{\left(\sqrt{2x^2 + 4} \right)^3}} \\
\end{array}$
Part (ii):
1. First we will split the given function
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{e^{\sec^2 x}~+~3 \cos^{-1} x} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{u+v} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{du}{dx}~+~\frac{dv}{dx}} \\
\end{array}$
2. Now we will find $\rm{\frac{du}{dx}}$:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{u} & {~=~} &{e^{\sec^2 x}} \\
{~\color{magenta} 2 } &{\implies} &{\log u} & {~=~} &{\sec^2 x.\log_e e} \\
{~\color{magenta} 3 } &{\implies} &{\log u} & {~=~} &{\sec^2 x} \\
{~\color{magenta} 4 } &{\implies} &{\frac{d}{dx} \left[\log u \right]} & {~=~} &{\frac{d}{dx} \left[\sec^2 x \right]} \\
{~\color{magenta} 5 } &{\implies} &{\frac{1}{u}.\frac{du}{dx}} & {~=~} &{2 \sec x . \sec x \tan x} \\
{~\color{magenta} 6 } &{\implies} &{\frac{du}{dx}} & {~=~} &{u. 2 \sec x . \sec x \tan x} \\
{~\color{magenta} 7 } &{{}} &{{}} & {~=~} &{2 e^{\sec^2 x} \sec^2 x \tan x} \\
\end{array}$
3. Now we will find $\rm{\frac{dv}{dx}}$:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{v} & {~=~} &{3 \cos^{-1} x} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dv}{dx}} & {~=~} &{3 \frac{d}{dx} \left[\cos^{-1} x \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{3. \frac{(-1)}{\sqrt{1 – x^2}}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{-3}{\sqrt{1 – x^2}}} \\
\end{array}$
4. So based on (1), we get:
$\rm{\frac{dy}{dx}~=~\frac{du}{dx}~+~\frac{dv}{dx}}$
= $\rm{2 e^{\sec^2 x} \sec^2 x \tan x~-~\frac{3}{\sqrt{1 – x^2}}}$
Part (iii):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\log_7 (\log x)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{\log_e (\log x)}{\log_e 7}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{1}{\log_e 7} \frac{d}{dx} \left[\log_e (\log x) \right]} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{\log_e 7} .\frac{1}{\log x}.\frac{1}{x}} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{\frac{1}{x \log 7 \log x}} \\
\end{array}$
◼ Remarks:
• 2 (magenta color): Here we apply change of base rule.
• 5 (magenta color): When the base of logarithm is not specified, it implies that, the base is e.
Solved example 21.67
Differentiate w.r.t x the following functions:
(i) $\rm{\cos^{-1} (\sin x)}$
(ii) $\rm{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)}$
(iii) $\rm{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}$
Solution:
Part (i):
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\cos^{-1} (\sin x)} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{-1}{\sqrt{1 - \sin^2 x}}.\frac{d}{dx} (\sin x)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{-1}{\sqrt{\cos^2 x}}.\cos x} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{-1} \\
\end{array}$
Alternate method:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\cos^{-1} (\sin x)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\cos^{-1} \left[\cos\left(\frac{\pi}{2} ~-~x \right) \right]} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{\pi}{2} ~-~x} \\
{~\color{magenta} 4 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{-1} \\
\end{array}$
Part (ii): $\rm{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)}$
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\tan^{-1} \left(\frac{\sin x}{1 + \cos x} \right)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\tan^{-1} \left(\tan \frac{x}{2} \right)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{x}{2}} \\
{~\color{magenta} 4 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{\frac{1}{2}} \\
\end{array}$
◼ Remarks:
• 2 (magenta color): Here we use identities:
$\rm{\cos x ~=~ 1 - 2 \sin^2 \frac{x}{2}}$
$\rm{\cos x ~=~ 2 \cos^2 \frac{x}{2} - 1}$
Combining these two identities, we get:
$\rm{\tan \frac{x}{2}~=~\frac{\sin x}{1 + \cos x}}$
Part (iii): $\rm{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)}$
1. First we will simplify the expression.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Let}~2^x} & {~=~} &{t} \\
{~\color{magenta} 2 } &{{}} &{\text{Given:}~\frac{2^{x+1}}{1+4^x}} & {~=~} &{\frac{2.2^x}{1+(2.2)^x}} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{2.2^x}{1+2^x .2^x}} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{2t}{1+t^2}} \\
\end{array}$
2. Next we will write 't' in terms of 𝜃.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\text{Let}~t} & {~=~} &{\tan \frac{\theta}{2}} \\
{~\color{magenta} 2 } &{{}} &{\text{Then:}~\frac{2t}{1+t^2}} & {~=~} &{\sin \theta} \\
\end{array}$
(Identity 15. List of identities can be seen here)
3. Next we will write 'y' in terms of 𝜃.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{\sin^{-1} \left(\frac{2^{x+1}}{1 + 4^x} \right)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\sin^{-1} \left(\frac{2t}{1 + t^2} \right)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\sin^{-1}(\sin \theta)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\theta} \\
\end{array}$
4. Next we will write 𝜃 in terms of 'x'.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{2^x} & {~=~} &{t} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\tan \frac{\theta}{2}} \\
{~\color{magenta} 3 } &{\implies} &{\frac{\theta}{2}} & {~=~} &{\tan^{-1}(2^x)} \\
{~\color{magenta} 4 } &{\implies} &{\theta} & {~=~} &{2. \tan^{-1}(2^x)} \\
\end{array}$
5. Based on (3), we can now connect 'y' and 'x'.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{y} & {~=~} &{2. tan^{-1}(2^x)} \\
{~\color{magenta} 2 } &{\implies} &{\frac{dy}{dx}} & {~=~} &{2. \frac{1}{1 + (2^x)^2}.\frac{d}{dx}(2^x)} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{2. \frac{1}{1 + (2^2)^x}.\frac{d}{dx}(2^x)} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{2}{1 + 4^x}.\frac{d}{dx}(2^x)} \\
\end{array}$
6. So our next task is to find the derivative of $\rm{2^x}$.
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{u} & {~=~} &{2^x} \\
{~\color{magenta} 2 } &{\implies} &{\log u} & {~=~} &{x \log 2} \\
{~\color{magenta} 3 } &{\implies} &{\frac{1}{u}.\frac{du}{dx}} & {~=~} &{1. \log 2} \\
{~\color{magenta} 4 } &{\implies} &{\frac{du}{dx}} & {~=~} &{u.\log 2} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{2^x \log 2} \\
\end{array}$
7. So from (5), we get:
$\begin{array}{ll} {~\color{magenta} 1 } &{{}} &{\frac{dy}{dx}} & {~=~} &{\frac{2}{1 + 4^x}.\frac{d}{dx}(2^x)} \\
{~\color{magenta} 2 } &{{}} &{{}} & {~=~} &{\frac{2}{1 + 4^x}.(2^x) \log 2} \\
{~\color{magenta} 3 } &{{}} &{{}} & {~=~} &{\frac{2^{x+1} \log 2}{1 + 4^x}} \\
\end{array}$
In the next section, we will see a few more solved examples.
Previous
Contents
Next
Copyright©2024 Higher secondary mathematics.blogspot.com
No comments:
Post a Comment