The value of [tex]\frac{d x}{d t}[/tex] is 20
Explanation:
Given that the expression is [tex]y=\sqrt{2 x+1}[/tex]
We need to determine the value of [tex]\frac{d x}{d t}[/tex] when [tex]x=12[/tex]
Value of [tex]\frac{d x}{d t}[/tex]:
Let us differentiate both sides of the equation with respect to t.
Thus, we have,
[tex]\frac{d y}{d t}=\frac{d(\sqrt{2 x+1})}{d t}[/tex]
Applying the chain rule, we get,
[tex]\frac{d y}{d t}=\frac{d(\sqrt{2 x+1})}{d x} \times \frac{d x}{d t}[/tex]
Differentiating, we get,
[tex]\frac{d y}{d t}=\frac{1}{2 \sqrt{2 x+1}} \times(2) \times \frac{d x}{d t}[/tex]
Simplifying, we get,
[tex]\frac{d y}{d t}=\frac{1}{\sqrt{2 x+1}} \times \frac{d x}{d t}[/tex]
Now, we shall determine the value of [tex]\frac{d x}{d t}[/tex] when [tex]x=12[/tex]
Substituting [tex]x=12[/tex] and [tex]\frac{dy}{dt}=4[/tex], we have,
[tex]4=\frac{1}{\sqrt{2(12)+1}} \times \frac{d x}{d t}[/tex]
Simplifying the values, we get,
[tex]4=\frac{1}{\sqrt{25}} \times \frac{d x}{d t}[/tex]
[tex]4=\frac{1}{5} \times \frac{d x}{d t}[/tex]
[tex]20=\frac{d x}{d t}[/tex]
Thus, the value of [tex]\frac{d x}{d t}[/tex] is 20
