Solved on Mar 08, 2024

Find the derivative of $y = \sqrt{8 + t^2}$ .

STEP 1

Assumptions
1. The function to differentiate is $y = \sqrt{8 + t^2}$ .
2. We will use the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

STEP 2

Identify the outer function and the inner function. In this case, the outer function is $f(u) = \sqrt{u}$ and the inner function is $g(t) = 8 + t^2$ .

STEP 3

Compute the derivative of the outer function with respect to its argument $u$ , which is $f'(u) = \frac{1}{2\sqrt{u}}$ .

STEP 4

Compute the derivative of the inner function with respect to $t$ , which is $g'(t) = 2t$ .

STEP 5

Apply the chain rule, which states that the derivative of the composite function $y = f(g(t))$ with respect to $t$ is $f'(g(t)) \cdot g'(t)$ .

STEP 6

Substitute the derivatives from STEP_3 and STEP_4 into the chain rule formula.
$\frac{\mathrm{d}y}{\mathrm{d}t} = f'(g(t)) \cdot g'(t) = \frac{1}{2\sqrt{g(t)}} \cdot 2t$

STEP 7

Replace $g(t)$ with $8 + t^2$ in the derivative of the outer function.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{2\sqrt{8 + t^2}} \cdot 2t$

STEP 8

Simplify the expression by canceling the 2 in the numerator and the denominator.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{2t}{2\sqrt{8 + t^2}}$

STEP 9

Further simplify the expression.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{t}{\sqrt{8 + t^2}}$

STEP 10

Write the final expression for the differential of $y$ with respect to $t$ .
$\mathrm{d}y = \frac{t}{\sqrt{8 + t^2}} \mathrm{d}t$
The differential of $y = \sqrt{8 + t^2}$ with respect to $t$ is $\mathrm{d}y = \frac{t}{\sqrt{8 + t^2}} \mathrm{d}t$ .

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Find the derivative of $y = \sqrt{8 + t^2}$ .

STEP 1

Assumptions
1. The function to differentiate is $y = \sqrt{8 + t^2}$ .
2. We will use the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.

STEP 2

Identify the outer function and the inner function. In this case, the outer function is $f(u) = \sqrt{u}$ and the inner function is $g(t) = 8 + t^2$ .

STEP 3

Compute the derivative of the outer function with respect to its argument $u$ , which is $f'(u) = \frac{1}{2\sqrt{u}}$ .

STEP 4

Compute the derivative of the inner function with respect to $t$ , which is $g'(t) = 2t$ .

STEP 5

Apply the chain rule, which states that the derivative of the composite function $y = f(g(t))$ with respect to $t$ is $f'(g(t)) \cdot g'(t)$ .

STEP 6

Substitute the derivatives from STEP_3 and STEP_4 into the chain rule formula.
$\frac{\mathrm{d}y}{\mathrm{d}t} = f'(g(t)) \cdot g'(t) = \frac{1}{2\sqrt{g(t)}} \cdot 2t$

STEP 7

Replace $g(t)$ with $8 + t^2$ in the derivative of the outer function.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{2\sqrt{8 + t^2}} \cdot 2t$

STEP 8

Simplify the expression by canceling the 2 in the numerator and the denominator.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{2t}{2\sqrt{8 + t^2}}$

STEP 9

Further simplify the expression.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{t}{\sqrt{8 + t^2}}$

STEP 10

Write the final expression for the differential of $y$ with respect to $t$ .
$\mathrm{d}y = \frac{t}{\sqrt{8 + t^2}} \mathrm{d}t$
The differential of $y = \sqrt{8 + t^2}$ with respect to $t$ is $\mathrm{d}y = \frac{t}{\sqrt{8 + t^2}} \mathrm{d}t$ .

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Find the derivative of y=8+t2y = \sqrt{8 + t^2}y=8+t2​.

STEP 1

STEP 2

Identify the outer function and the inner function. In this case, the outer function is f(u)=u f(u) = \sqrt{u} f(u)=u​ and the inner function is g(t)=8+t2 g(t) = 8 + t^2 g(t)=8+t2.

STEP 3

Compute the derivative of the outer function with respect to its argument u u u, which is f′(u)=12u f'(u) = \frac{1}{2\sqrt{u}} f′(u)=2u​1​.

STEP 4

Compute the derivative of the inner function with respect to t t t, which is g′(t)=2t g'(t) = 2t g′(t)=2t.

STEP 5

Apply the chain rule, which states that the derivative of the composite function y=f(g(t)) y = f(g(t)) y=f(g(t)) with respect to t t t is f′(g(t))⋅g′(t) f'(g(t)) \cdot g'(t) f′(g(t))⋅g′(t).

STEP 6

Substitute the derivatives from STEP_3 and STEP_4 into the chain rule formula.dydt=f′(g(t))⋅g′(t)=12g(t)⋅2t \frac{\mathrm{d}y}{\mathrm{d}t} = f'(g(t)) \cdot g'(t) = \frac{1}{2\sqrt{g(t)}} \cdot 2t dtdy​=f′(g(t))⋅g′(t)=2g(t)​1​⋅2t

STEP 7

Replace g(t) g(t) g(t) with 8+t2 8 + t^2 8+t2 in the derivative of the outer function.dydt=128+t2⋅2t \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{2\sqrt{8 + t^2}} \cdot 2t dtdy​=28+t2​1​⋅2t

STEP 8

Simplify the expression by canceling the 2 in the numerator and the denominator.dydt=2t28+t2 \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{2t}{2\sqrt{8 + t^2}} dtdy​=28+t2​2t​

STEP 9

Further simplify the expression.dydt=t8+t2 \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{t}{\sqrt{8 + t^2}} dtdy​=8+t2​t​

STEP 10

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Find the derivative of $y = \sqrt{8 + t^2}$ .

Identify the outer function and the inner function. In this case, the outer function is $f(u) = \sqrt{u}$ and the inner function is $g(t) = 8 + t^2$ .

Compute the derivative of the outer function with respect to its argument $u$ , which is $f'(u) = \frac{1}{2\sqrt{u}}$ .

Compute the derivative of the inner function with respect to $t$ , which is $g'(t) = 2t$ .

Apply the chain rule, which states that the derivative of the composite function $y = f(g(t))$ with respect to $t$ is $f'(g(t)) \cdot g'(t)$ .

Substitute the derivatives from STEP_3 and STEP_4 into the chain rule formula.
$\frac{\mathrm{d}y}{\mathrm{d}t} = f'(g(t)) \cdot g'(t) = \frac{1}{2\sqrt{g(t)}} \cdot 2t$

Replace $g(t)$ with $8 + t^2$ in the derivative of the outer function.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{1}{2\sqrt{8 + t^2}} \cdot 2t$

Simplify the expression by canceling the 2 in the numerator and the denominator.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{2t}{2\sqrt{8 + t^2}}$

Further simplify the expression.
$\frac{\mathrm{d}y}{\mathrm{d}t} = \frac{t}{\sqrt{8 + t^2}}$