Question

If $t=\sqrt{x}+4$, then ${\left(\frac{dx}{dt}\right)}_{t=4}$ is:

• A. 4
• B. Zero
• C. 8
• D. 16

Answer 1

Answer: (B)

$t=\sqrt{x}+4$
$\Rightarrow x=(t-4{)}^2=t^2-8t+16$
$\Rightarrow \frac{dx}{dt}=2t-8$
${\Rightarrow \frac{dx}{dt}|}_{t=4}=2\times 4-8=0$