How do you find the derivative of #(3x+1)^(3/2) (2x+4)#?
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To find the derivative of the function ( (3x+1)^{\frac{3}{2}} \times (2x+4) ), you can use the product rule combined with the chain rule. The product rule states that if you have two functions ( u(x) ) and ( v(x) ), the derivative of their product is given by ( (u'v + uv') ). The chain rule states that if you have a composite function ( f(g(x)) ), its derivative is given by ( f'(g(x)) \times g'(x) ). Applying these rules:
Let ( u(x) = (3x+1)^{\frac{3}{2}} ) and ( v(x) = (2x+4) ). Now, find the derivatives of ( u(x) ) and ( v(x) ) using the chain rule and the power rule:
[ u'(x) = \frac{3}{2}(3x+1)^{\frac{1}{2}} \times 3 ] [ v'(x) = 2 ]
Now apply the product rule:
[ (3x+1)^{\frac{3}{2}} \times 2 + (3x+1)^{\frac{1}{2}} \times (2x+4) \times 3 ]
Simplify this expression to get the derivative of the function.
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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.
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.
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.
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.
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