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

Studdy Solution

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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