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Exponential derivatives Quiz 2

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 If $F={{e}^{t+\tan t}}$, write $\dfrac{dF}{dt}$ at $t=0$.

 Find derivative using chain rule. Then substitute $t=0$ in it.

  $\text{Consider }F={{e}^{t+\tan t}} \\ \text{Differentiate both sides with respect to “$t$”} \\ \dfrac{dF}{dt}={{e}^{t+\tan t}}\times \dfrac{d}{dt}\left( t+\tan t \right) \\ \Rightarrow {{e}^{t+\tan t}}\left( 1+{{\sec }^{2}}t \right) \\ \text{Substitute }t=0 \\ {{\left[ \dfrac{dF}{dt} \right]}_{t=0}}={{e}^{0}}\left( 1+{{\sec }^{2}}0 \right)=\left( 1+1 \right)=2$

 Evaluate the derivative of function $f\left( x \right)={{\tan }^{-1}}\left( {{e}^{4x+5}} \right)$ with respect to $'x'$. 

  $\text{Consider the function }f\left( x \right)={{\tan }^{-1}}\left( {{e}^{4x+5}} \right) \\ \text{Differentiate both side with respect to $'x'$} \\ f'\left( x \right)=\dfrac{1}{1+{{\left( {{e}^{4x+5}} \right)}^{2}}}\times \dfrac{d}{dx}\left( {{e}^{4x+5}} \right) \\ \text{Solve further, } \\ f'\left( x \right)=\dfrac{1}{1+{{e}^{8x+10}}}\left[ {{e}^{4x+5}}\times 4 \right] \\ f'\left( x \right)=\dfrac{4{{e}^{4x+5}}}{1+{{e}^{8x+10}}}$

 Let $y={{e}^{\tan x}}$and $z={{e}^{\cos x}}$. Find the value of $A+B$, where $A=\dfrac{dy}{dx}$ at $x=0$ and $B=\dfrac{dz}{dx}$ at $x=0$.

 Find derivative of $y$ and $z$ using chain rule. Then substitute $x=0$ in both of them.

  $\text{Consider }y={{e}^{\operatorname{tanx}}} \\ \text{Then, }\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{e}^{\tan x}} \right) \\ \Rightarrow {{e}^{\tan x}}.{{\sec }^{2}}x \\ \text{Substitute }x=0 \\ \therefore {{\left[ \dfrac{dy}{dx} \right]}_{x=0}}={{e}^{\tan 0}}\times {{\sec }^{2}}0={{e}^{0}}\times 1=1 \\ \text{Consider, }z={{e}^{\cos x}} \\ \text{Then, }\dfrac{dz}{dx}=-{{e}^{\cos x}}.\sin x \\ \text{Substitute }x=0 \\ \therefore {{\left[ \dfrac{dz}{dx} \right]}_{x=0}}=-{{e}^{\cos 0}}\times \sin {{0}^{o}}=-e\times 0=0 \\ A+B=1+0=1$

 If $y={{e}^{{{\sec }^{-1}}\left( \dfrac{7\sqrt{x}+5}{7\sqrt{x}-5} \right)+{{\sin }^{-1}}\left( \dfrac{7\sqrt{x}-5}{7\sqrt{x}+5} \right)}}$ then find $\dfrac{dy}{dx}$.

 Use trigonometric properties to simplify the given function and then find derivatives.

  $\text{Consider the given function} \\ y={{e}^{{{\sec }^{-1}}\left( \dfrac{7\sqrt{x}+5}{7\sqrt{x}-5} \right)+{{\sin }^{-1}}\left( \dfrac{7\sqrt{x}-5}{7\sqrt{x}+5} \right)}} ----(1) \\ {{\sin }^{-1}}\left( \dfrac{7\sqrt{x}-5}{7\sqrt{x}+5} \right)=\cos e{{c}^{-1}}\left( \dfrac{7\sqrt{x}+5}{7\sqrt{x}-5} \right) ----(2) \left[\text{Using }{{\sin }^{-1}}x={{\operatorname{cosec}}^{-1}}\left(\dfrac{1}{x}\right)\right] \\ \text{Substitute equation (ii) in equation (i)} \\ y={{e}^{{{\sec }^{-1}}\left( \dfrac{7\sqrt{x}+5}{7\sqrt{x}-5} \right)+\cos e{{c}^{-1}}\left( \dfrac{7\sqrt{x}+5}{7\sqrt{x}-5} \right)}} \\ \Rightarrow y={{e}^{\frac{\pi }{2}}} \,\,\,\,\,\,	(\text{Using }{{\sec }^{-1}}x+{{\operatorname{cosec}}^{-1}}x=\dfrac{\pi}{2}) \\ \Rightarrow \dfrac{dy}{dx}=0$

 For the equation ${{e}^{t}}+tz=0$, find the value of $\dfrac{{{d}^{2}}z}{d{{t}^{2}}}$at $t=1$.

 Differentiate both sides of the given equation twice and substitute $t=1$ and $z=-e$. 

  $\text{Consider the equation }{{e}^{t}}+tz=0 \\ \text{Differentiate both sides with respect to $'t'$.} \\ {{e}^{t}}+t\dfrac{dz}{dt}+z.1=0 \\ \Rightarrow \dfrac{dz}{dt}=\dfrac{-{{e}^{t}}-z}{t} ----(i) \\ \text{Again differentiate with respect to $'t'$} \\ \dfrac{{{d}^{2}}z}{d{{t}^{2}}}=\dfrac{t\times \dfrac{d}{dt}\left( -{{e}^{t}}-z \right)-\left( -{{e}^{t}}-z \right)\times \dfrac{d}{dt}\left( t \right)}{{{t}^{2}}} \\ \Rightarrow \dfrac{t\left( -{{e}^{t}}-\dfrac{dz}{dt} \right)+\left( {{e}^{t}}+z \right)1}{{{t}^{2}}} ----(ii) \\ \text{Substitute }\dfrac{dz}{dt} \text{ from equation (i) in equation (ii)} \\ \dfrac{{{d}^{2}}z}{d{{t}^{2}}}=\dfrac{t\left[ -{{e}^{t}}+\dfrac{{{e}^{t}}+z}{t} \right]+{{e}^{t}}+z}{{{t}^{2}}} \\ \text{Also, at }t=1, z=-e \\ \Rightarrow \dfrac{{{d}^{2}}z}{d{{t}^{2}}}=\dfrac{-{{t}^{2}}{{e}^{t}}+2t{{e}^{t}}+2tz}{{{t}^{3}}} \\ \Rightarrow \dfrac{-t{{e}^{t}}+2{{e}^{t}}+2z}{{{t}^{2}}} \\ \text{Substitute  }t=1\text{ and }z=-e \\ \dfrac{{{d}^{2}}z}{d{{t}^{2}}}=\dfrac{e-2e}{{{(1)}^{2}}}=-e$