   Chapter 4.1, Problem 81E

Chapter
Section
Textbook Problem

PNTAM EXAM CHALLEGSuppose of f and g are non-constant, differentiable, real valued functions defined on ( − ∞ ,   ∞ ) . Furthermore, suppose that for each pair of real numbers x and y, f ( x + y ) = f ( x ) f ( y ) − g ( x ) g ( y )             and g ( x , y) =  f ( x ) g ( y ) +  g ( x ) f ( y ) If f ' ( 0 ) = 0 , prove that ( f ( x ) ) 2 + ( g ( x ) ) 2 = 1 for all x.

To determine

To Calculate: dydx using logarithmic differentiation.

Explanation

Given: y=x(x1)32x+1 and x>1

Formula Used:

Calculation:

Consider the given equation is:

y=x(x1)32x+1

Taking logarithmic on both sides, we get

lny=lnx(x1)32x+1

Using the property of logarithm:

ln(ab)=lnalnb,ln(ab)=lna+lnb and lnan=nlna

lny=lnx(x1)32lnx+1=lnx+32ln(x1)ln(x+1)12=lnx+32ln(x1)12ln(x+1)

Differentiating with respect to x, we get

1y.dydx=1x+321x112

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