If u = e^{ax} bx and v = e^{ax} bx, then what is udu/dx+vdv/dx equal to?

Correct Answer is : ae2axa e²ax}ae2ax

Correct Answer is : ae2axa e²ax}ae2ax

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UPSC NDA II 2018 Mathematics Solved Paper

Question : 87 of 120

Marks: +1, -0

u = e^{ax} \sin bx

v = e^{ax} \cos bx

u\frac{du}{dx}+v\frac{dv}{dx}

equal to?

Solution:

Concept:

Differentiate

u

and

v

using the product rule and chain rule, then combine the terms.

Explanation:

Given

u = e^{ax} \sin bx

and

v = e^{ax} \cos bx

Differentiate

u

with respect to

x

\frac{du}{dx} = a e^{ax} \sin bx + e^{ax} \cdot b \cos bx = e^{ax} (a \sin bx + b \cos bx)

Multiply

u

\frac{du}{dx}

u \frac{du}{dx} = e^{ax} \sin bx \cdot e^{ax} (a \sin bx + b \cos bx) = e^{2ax} (a \sin^2 bx + b \sin bx \cos bx)

Differentiate

v

with respect to

x

\frac{dv}{dx} = a e^{ax} \cos bx - e^{ax} \cdot b \sin bx = e^{ax} (a \cos bx - b \sin bx)

Multiply

v

\frac{dv}{dx}

v \frac{dv}{dx} = e^{ax} \cos bx \cdot e^{ax} (a \cos bx - b \sin bx) = e^{2ax} (a \cos^2 bx - b \sin bx \cos bx)

Add the two expressions:

u \frac{du}{dx} + v \frac{dv}{dx} = e^{2ax} (a \sin^2 bx + b \sin bx \cos bx + a \cos^2 bx - b \sin bx \cos bx)

The terms

b \sin bx \cos bx

cancel. Using

\sin^2 bx + \cos^2 bx = 1

, we get:

= e^{2ax} \cdot a \cdot (\sin^2 bx + \cos^2 bx) = a e^{2ax}

Answer:

a e^{2ax}

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