The velocity of a moving vehicle is given by the equation "\ud835\udc63 = (2\ud835\udc61 + 3)^4". Use the Chain Rule to determine an equation for the acceleration when π = ππ£ ππ‘ .
Given: "v=(2t+3)^4"
Require to find "a=\\frac{dv}{dt}" using Chain Rule.
Recollect the following (Chain Rule):
If "y=f(u)" and "u=g(x)" , then "\\frac{dy}{dx}=\\frac{dy}{du}\\frac{du}{dx}"
Let "u=(2t+3)"
Then "v=u^4,u=2t+3"
"\\Rightarrow \\frac{dv}{du}=4u^3,\\frac{du}{dt}=2(1)+0=2"
Using Chain Rule, we get
"\\frac{dv}{dt}=4u^3(2)=8u^3"
Substituting "u=2t+3," we get
"\\frac{dv}{dt}=8(2t+3)^3"
Therefore the acceleration "a=\\frac{dv}{dt}=8(2t+3)^3"
Comments
Derivative of the function u(t) is "\\frac{d}{dt} u(t)=(2t+3)'=2+0=2,"as derivative of t and constant is equal 1 and 0 respectively.
, dt du β =2(1)+0=2 where did this come from?
Dear Muhammad farooq, a solution of this question was already published. A new problem can be submitted as a new question or a new order.
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