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How do you differentiate #sqrtt*(1-t^2)#?

Answer 1

Isaiah Allen

It's quickest to multiply #\sqrt{t}# through with the distributive property and then use linearity and the power rule:

#d/dt(\sqrt{t}\cdot (1-t^{2}))=d/dt(t^{1/2}-t^{5/2})=\frac{1}{2}t^{-1/2}-\frac{5}{2}t^{3/2}.#

For extra fun, you can also use the product rule along with linearity and the power rule:

#d/dt(\sqrt{t}\cdot (1-t^{2}))=\frac{1}{2}t^{-1/2}(1-t^{2})+t^{1/2}\cdot (-2t)#

#=\frac{1}{2}t^{-1/2}-\frac{1}{2}t^{3/2}-2t^{3/2}=\frac{1}{2}t^{-1/2}-\frac{5}{2}t^{3/2}.#

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Answer 2

Elijah Abram

To differentiate ( \sqrt{t} \cdot (1 - t^2) ), you can use the product rule.

Let ( u = \sqrt{t} ) and ( v = 1 - t^2 ).
Differentiate ( u ) with respect to ( t ): ( \frac{du}{dt} = \frac{1}{2\sqrt{t}} ).
Differentiate ( v ) with respect to ( t ): ( \frac{dv}{dt} = -2t ).
Apply the product rule: ( \frac{d}{dt} (uv) = u\frac{dv}{dt} + v\frac{du}{dt} ).
Substitute ( u ), ( v ), ( \frac{du}{dt} ), and ( \frac{dv}{dt} ) into the product rule equation.

( \frac{d}{dt} (\sqrt{t} \cdot (1 - t^2)) = (\sqrt{t})(-2t) + (1 - t^2)\frac{1}{2\sqrt{t}} ).
Simplify if necessary.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.