If the value of $y = \log \left[ {\tan x} \right],$ find$\frac{{dy}}{{dx}}$.
Answer
363.9k+ views
Hint: you can differentiate it by using formulae of logarithm or
Using substitution method.
We have to differentiate it with respect to $x$.
We will use a substitution method.
Let, $\tan x = t$
$\begin{gathered}
\therefore y = \log t \\
dy = \frac{1}{t}dt \\
\end{gathered} $
$\frac{{dy}}{{dt}} = \frac{1}{t} = \frac{1}{{\tan x}}$$ \ldots \ldots \left( 1 \right)$
Now
$\begin{gathered}
t = \tan x \\
dt = {\sec ^2}xdx \\
\frac{{dt}}{{dx}} = {\sec ^2}x \ldots \ldots \left( 2 \right) \\
\end{gathered} $
On multiplying equation $\left( 1 \right)$and $\left( 2 \right)$
We get,
$\begin{gathered}
\frac{{dy}}{{dt}} \times \frac{{dt}}{{dx}} = \frac{{{{\sec }^2}x}}{{\tan x}} \\
\frac{{dy}}{{dx}} = \frac{{\cos x}}{{\cos x.\cos x.\sin x}} = \frac{1}{{\sin x.\cos x}} \\
\frac{{dy}}{{dx}} = \frac{1}{{\sin x.\cos x}} \\
\end{gathered} $
Is the required answer.
Note:Using a substitution method you can solve easily. You have to find
Differentiation with respect to $x$, then you can differentiate it with respect
to $t$ and further differentiate $t$ with respect to $x$
Using substitution method.
We have to differentiate it with respect to $x$.
We will use a substitution method.
Let, $\tan x = t$
$\begin{gathered}
\therefore y = \log t \\
dy = \frac{1}{t}dt \\
\end{gathered} $
$\frac{{dy}}{{dt}} = \frac{1}{t} = \frac{1}{{\tan x}}$$ \ldots \ldots \left( 1 \right)$
Now
$\begin{gathered}
t = \tan x \\
dt = {\sec ^2}xdx \\
\frac{{dt}}{{dx}} = {\sec ^2}x \ldots \ldots \left( 2 \right) \\
\end{gathered} $
On multiplying equation $\left( 1 \right)$and $\left( 2 \right)$
We get,
$\begin{gathered}
\frac{{dy}}{{dt}} \times \frac{{dt}}{{dx}} = \frac{{{{\sec }^2}x}}{{\tan x}} \\
\frac{{dy}}{{dx}} = \frac{{\cos x}}{{\cos x.\cos x.\sin x}} = \frac{1}{{\sin x.\cos x}} \\
\frac{{dy}}{{dx}} = \frac{1}{{\sin x.\cos x}} \\
\end{gathered} $
Is the required answer.
Note:Using a substitution method you can solve easily. You have to find
Differentiation with respect to $x$, then you can differentiate it with respect
to $t$ and further differentiate $t$ with respect to $x$
Last updated date: 25th Sep 2023
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