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]

W. r. to. x. is

(A) -29/15



(B) 7/3

(C) 31/15


(D) 29/15
​

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

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

$\displaystyle \sf{ \: }f(1)=3 \: , \: f'(1)= - \frac{1}{3} \: , \: g(1)= -4 \: , \: g'(1)=- \frac{8}{3}$

TO DETERMINE

The derivative of

$\sf{ \sqrt{ {[ \: f(x) \: ] }^{2} + {{[ \: g(x) \: ] }}^{2} \: } \: \: }$

CALCULATION

It is given that

$\displaystyle \sf{ \: }f(1)=3 \: , \: f'(1)= - \frac{1}{3} \: , \: g(1)= -4 \: , \: g'(1)=- \frac{8}{3}$

$\sf{ Let \: \: \: y = \sqrt{ {[ \: f(x) \: ] }^{2} + {{[ \: g(x) \: ] }}^{2} \: } \: \: } \: \: \: ......(1)$

For x = 1

$\sf{y = \sqrt{ {[ \: f(1) \: ] }^{2} + {{[ \: g(1) \: ] }}^{2} \: } \: \: }$

$\sf{ = \sqrt{9 + 16} \: \: }$

$= \sf{ \sqrt{25} \: }$

$\sf{ = 5 }$

From Equation (1)

$\sf{{y}^{2} = {[ \: f(x) \: ] }^{2} + {{[ \: g(x) \: ] }}^{2} \: \: \: }$

Differentiating both sides with respect to x we get

$\displaystyle \sf{ 2y \frac{dy}{dx} = 2f(x)f' (x) + 2g(x)g' (x) \: }$

$\implies \: \displaystyle \sf{ y \frac{dy}{dx} = f(x)f' (x) + g(x)g' (x) \: }$

Putting x = 1 in both sides

$\displaystyle \sf{ 5 \frac{dy}{dx} = f(1)f' (1) + g(1)g' (1) \: }$

$\implies \: \displaystyle \sf{ 5 \frac{dy}{dx} =3 \times ( - \frac{1}{3} ) + ( - 4) \times ( - \frac{8}{3} )\: }$

$\implies \: \displaystyle \sf{ 5 \frac{dy}{dx} = - 1 + \frac{32}{3} \: }$

$\implies \: \displaystyle \sf{ 5 \frac{dy}{dx} = \frac{29}{3} \: }$

$\implies \: \displaystyle \sf{ \frac{dy}{dx} = \frac{29}{15} \: }$

RESULT

$\displaystyle \sf{ The \: required \: derivative = \frac{29}{15} \: }$

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