Question

Let f(1)=3,f'(1)=-1/3,g(1)=-4 and g'(1)= -8/3 . The derivative of √[f(x)]^2+[g(x)^2]

Answer 1

Given :  f(1) = 3,f' (1)=- 1/3,g (1)=-4 and g' (1) =-8/3

To Find : The derivative of √[[f(x)]^2+ [g (x)]^2]  wrt x at x = 1

Solution:

derivative of √[[f(x)]²+ [g (x)]²]

= {  1/2√[[f(x)]²+ [g (x)]²]  }  { 2 f(x) f'(x)  + 2g(x) g'(x) }

at  x = 1

= {  1/2√[[f(1)]²+ [g (1)]²]  }  { 2 f(1) f'(1)  + 2g(1) g'(1) }

=  {  1/2√[[3]²+ [-4]²]  }  { 2 (3) (-1/3)  + 2(-4) (-8/3) }

=  {  1/2(5) }  { -2 + 64/3) }

=  { 1/10 } { 58/3}

= 58/30

= 29/15

29/15 is correct answer

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Answer 2

★ Solution:

derivative of √[[f(x)]²+ [g (x)]²]

= {  1/2√[[f(x)]²+ [g (x)]²]  }  { 2 f(x) f'(x)  + 2g(x) g'(x) }

at  x = 1

= {  1/2√[[f(1)]²+ [g (1)]²]  }  { 2 f(1) f'(1)  + 2g(1) g'(1) }

=  {  1/2√[[3]²+ [-4]²]  }  { 2 (3) (-1/3)  + 2(-4) (-8/3) }

=  {  1/2(5) }  { -2 + 64/3) }

=  { 1/10 } { 58/3}

= 58/30

= 29/15

29/15 is correct answer