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Mathematics Question on Second Order Derivative

The derivative of eaxcos⁡ bxe^{ax} \cos\,bx with respect to xx is reaxre^{ax} cos⁡ (bx−Tan−1ba)\cos\, (bx-Tan^{-1} \frac {b}{a}) When a>0,b>0a >0, b > 0,the value of rr is ............

A

abab

B

a+ba+b

C

a2+b2\sqrt {a^2+b^2}

D

1ab\frac {1} {\sqrt {ab}}

Answer

a2+b2\sqrt {a^2+b^2}

Explanation

Solution

Given, ddx(eaxcos⁡bx)=reaxcos⁡(bx+α)\frac{d}{d x}\left(e^{a x} \cos b x\right)=r e^{a x} \cos (b x+\alpha), then
r=?r=?
Let y=eax⋅cos⁡bxy=e^{a x} \cdot \cos b x
Let y=eax⋅cos⁡bxy=e^{a x} \cdot \cos b x
dydx=aeax⋅cos⁡bx−beax⋅sin⁡bx\frac{d y}{d x}=a e^{a x} \cdot \cos b x-b e^{a x} \cdot \sin b x
dydx=eax(acos⁡bx−bsin⁡bx)\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, dydx=eax⋅rcos⁡bx⋅cos⁡α−sin⁡bx⋅sin⁡α\frac{d y}{d x} =e^{a x} \cdot r\\{\cos b x \cdot \cos \alpha-\sin b x \cdot \sin \alpha\\}
dydx=eax⋅rcos⁡(bx+α)\frac{d y}{d x}=e^{a x} \cdot r \cos (b x+\alpha)...(ii)
Where, tan⁡α=ba⇒α=tan⁡−1(ba)\tan \alpha=\frac{b}{a} \Rightarrow \alpha=\tan ^{-1}\left(\frac{b}{a}\right)
and r2(cos⁡2α+sin⁡2α)=a2+b2r^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)=a^{2}+b^{2}
r2=a2+b2r^{2}=a^{2}+b^{2}
r=a2+b2r=\sqrt{a^{2}+b^{2}}