The rate of change of ( x^ 2 + 16 ) with respect to (x)/(x - 1) at x = 3 is

Correct answer: C. - (12)/(5) Let y = x^ 2 + 16 andz = (x)/(x - 1) On differentiating w.r.t., x , we get, (dz)/(dx) = (x - 1 - x)/((x - 1)2) = (- 1)/( (x - 1)2) and (dy)/(dx) = 1 × 2x 2 x^ 2 + 16 ⇒ (dy)/(dz) = x x^ 2 + 16 - 1 array ll (x - 1)2 array At x = 3 (dy)/(dz) = ((3)/(5))(( 4 ))/(( - 1 )) = (- 12)/(5) &

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Differentiation

NTA Abhyas 2020 · Mathematics · Differentiation

The rate of change of ( x^ 2 + 16 ) with respect to (x)/(x - 1) at...

The rate of change of

\sqrt{( x^{2} + 16 )} with respect to \dfrac{x}{x - 1}

x = 3

Options

A $2$
B $\dfrac{11}{5}$
C $- \dfrac{12}{5}$
D $- 3$

Correct answer

C. $- \dfrac{12}{5}$

Step-by-step solution

Let

y = \sqrt{x^{2} + 16}andz = \dfrac{x}{x - 1}

On differentiating w.r.t.,

x

, we get,

\dfrac{dz}{dx} = \dfrac{x - 1 - x}{\left(x - 1\right)2}

= \dfrac{- 1}{

\left(x - 1\right)2}

and

\dfrac{dy}{dx} = \dfrac{1 \times 2x}{2\sqrt{x^{2} + 16}}

⇒ \dfrac{dy}{dz} = \dfrac{x\sqrt{x^{2} + 16}}{- 1

\begin{array}{ll}\left(x - 1\right)2\end{array}

}

x = 3

\dfrac{dy}{dz} = \left(\dfrac{3}{5}\right)\dfrac{( 4 )}{( - 1 )} = \dfrac{- 12}{5}

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