Question

Differentiate with respect to $x$:
$
  {\text{(a) tan}}h4x \\
  {\text{(b) sec}}h2x \\
$

Answer 1

Hint: In this question first we have to find hyperbolic functions ${\text{tan}}h4x$ and ${\text{sec}}h2x$ in terms of ${e^{kx}}$ by using relations $\sin hkx = \dfrac{{{e^{kx}} - {e^{ - kx}}}}{2}$ and $\cos hkx = \dfrac{{{e^{kx}} + {e^{ - kx}}}}{2}$. Then we differentiate ${\text{tan}}h4x$ and ${\text{sec}}h2x$ w.r.t. $x$.

Complete Step-by-Step solution:
Hyperbolic functions
$
   \Rightarrow \sin hkx = \dfrac{{{e^{kx}} - {e^{ - kx}}}}{2}{\text{ eq}}{\text{.1}} \\
   \Rightarrow \cos hkx = \dfrac{{{e^{kx}} + {e^{ - kx}}}}{2}{\text{ eq}}{\text{.2}} \\
 $
Differentiation of $\dfrac{u}{v}$ w.r.t. to$x$
$ \Rightarrow \dfrac{d}{{dx}}\dfrac{u}{v} = \dfrac{{\dfrac{{vdu}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}{\text{ eq}}{\text{.3}}$
${\text{(a) tan}}h4x$
    Divide eq.1 and eq.2, we get
 $ \Rightarrow \tanh kx{\text{ }} = {\text{ }}\dfrac{{{e^{kx}} - {e^{ - kx}}}}{{{e^{kx}} + {e^{ - kx}}}}$
Put h=4 in above equation we get
$ \Rightarrow \tan 4kx{\text{ }} = {\text{ }}\dfrac{{{e^{4x}} - {e^{ - 4x}}}}{{{e^{4x}} + {e^{ - 4x}}}}{\text{ eq}}{\text{.4}}$
Differentiate eq.4 w.r.t to $x$using formula of eq,3
$$$$$
   \Rightarrow \dfrac{{d\tan h4x}}{{dx}} = {\text{ }}\dfrac{d}{{dx}}(\dfrac{{{e^{4x}} - {e^{ - 4x}}}}{{{e^{4x}} + {e^{ - 4x}}}}) \\
  {\text{ = }}\dfrac{{({e^{4x}} + {e^{ - 4x}})\dfrac{d}{{dx}}({e^{4x}} - {e^{ - 4x}}) - ({e^{4x}} - {e^{ - 4x}})\dfrac{d}{{dx}}({e^{4x}} + {e^{ - 4x}})}}{{{{({e^{4x}} + {e^{ - 4x}})}^2}}} \\
  {\text{ = }}\dfrac{{4{{({e^{4x}} + {e^{ - 4x}})}^2} - 4{{({e^{4x}} - {e^{ - 4x}})}^2}}}{{{{({e^{4x}} + {e^{ - 4x}})}^2}}} \\
  {\text{ = 4\{ 1}} - {{\text{(}}\dfrac{{{e^{4x}} - {e^{ - 4x}}}}{{{e^{4x}} + {e^{ - 4x}}}})^2}\} \\
  {\text{ = 4(1}} - \tan {h^2}4x) \\
 $
${\text{(b) sec}}h2x$
We know that $\sec h4x = \dfrac{1}{{\cos h4x}}$
$ \Rightarrow \sec h4x = \dfrac{2}{{{e^{kx}} + {e^{ - kx}}}}$ eq.5
Differentiation eq. 4 w.r.t to $x$using formula of eq,3
$
   \Rightarrow \dfrac{d}{{dx}}\sec h2x = \dfrac{d}{{dx}}(\dfrac{2}{{{e^{2x}} + {e^{ - 2x}}}}) \\
  {\text{ = }}\dfrac{{({e^{2x}} + {e^{ - 2x}})\dfrac{d}{{dx}}2 - 2\dfrac{d}{{dx}}({e^{2x}} + {e^{ - 2x}})}}{{{{({e^{2x}} + {e^{ - 2x}})}^2}}} \\
  {\text{ = }}\dfrac{{0 - 4({e^{2x}} + {e^{ - 2x}})}}{{{{({e^{2x}} + {e^{ - 2x}})}^2}}} \\
  {\text{ = }} - 4\dfrac{1}{{({e^{2x}} + {e^{ - 2x}})}}.\dfrac{{({e^{2x}} - {e^{ - 2x}})}}{{({e^{2x}} + {e^{ - 2x}})}} \\
  {\text{ = }} - 4\sec h2x.\tan h2x \\
 $

Note: Whenever you get this type of question the key concept to solve is that use the hyperbolic functions $\sin hkx = \dfrac{{{e^{kx}} - {e^{ - kx}}}}{2}$and $\cos hkx = \dfrac{{{e^{kx}} + {e^{ - kx}}}}{2}$ to get other relations by just applying simple operations like dividing, adding etc. Remember one thing that hyperbolic functions are different from trigonometric functions.