Question:

The derivative of $e^{ax} \cos\,bx$ with respect to $x$ is $re^{ax}$ $\cos\, (bx-Tan^{-1} \frac {b}{a})$ When $a >0, b > 0$,the value of $r$ is ............

Updated On: May 19, 2024
  • $ab$
  • $a+b$
  • $\sqrt {a^2+b^2}$
  • $\frac {1} {\sqrt {ab}}$
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The Correct Option is C

Solution and Explanation

Given, $\frac{d}{d x}\left(e^{a x} \cos b x\right)=r e^{a x} \cos (b x+\alpha)$, then
$r=?$
Let $y=e^{a x} \cdot \cos b x$
Let $y=e^{a x} \cdot \cos b x$
$\frac{d y}{d x}=a e^{a x} \cdot \cos b x-b e^{a x} \cdot \sin b x$
$\frac{d y}{d x}=e^{a x}(a \cos b x-b \sin b x)$
$a=r \cos \alpha
b=r \sin \alpha\}$...(i)
Then, $\frac{d y}{d x} =e^{a x} \cdot r\{\cos b x \cdot \cos \alpha-\sin b x \cdot \sin \alpha\}$
$\frac{d y}{d x}=e^{a x} \cdot r \cos (b x+\alpha)$...(ii)
Where, $\tan \alpha=\frac{b}{a} \Rightarrow \alpha=\tan ^{-1}\left(\frac{b}{a}\right)$
and $r^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=a^{2}+b^{2}$
$r^{2}=a^{2}+b^{2}$
$r=\sqrt{a^{2}+b^{2}}$
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Concepts Used:

Second-Order Derivative

The Second-Order Derivative is the derivative of the first-order derivative of the stated (given) function. For instance, acceleration is the second-order derivative of the distance covered with regard to time and tells us the rate of change of velocity. 

As well as the first-order derivative tells us about the slope of the tangent line to the graph of the given function, the second-order derivative explains the shape of the graph and its concavity. 

The second-order derivative is shown using \(f’’(x)\text{ or }\frac{d^2y}{dx^2}\).