Solved on Mar 08, 2024

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

STEP 1

Assumptions
1. The function to differentiate is y=8+t2 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)=u f(u) = \sqrt{u} and the inner function is g(t)=8+t2 g(t) = 8 + t^2 .

STEP 3

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

STEP 4

Compute the derivative of the inner function with respect to t t , which is 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)) with respect to t t is f(g(t))g(t) f'(g(t)) \cdot 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

STEP 7

Replace g(t) g(t) with 8+t2 8 + t^2 in the derivative of the outer function.
dydt=128+t22t \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.
dydt=2t28+t2 \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{2t}{2\sqrt{8 + t^2}}

STEP 9

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

STEP 10

Write the final expression for the differential of y y with respect to t t .
dy=t8+t2dt \mathrm{d}y = \frac{t}{\sqrt{8 + t^2}} \mathrm{d}t
The differential of y=8+t2 y = \sqrt{8 + t^2} with respect to t t is dy=t8+t2dt \mathrm{d}y = \frac{t}{\sqrt{8 + t^2}} \mathrm{d}t .

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