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A car is travelling with a velocity of "0.5t^2+t+2" m/s at t=0 (where t is in seconds), find the acceleration of the car at a) t=0 b)t=2

Acceleration can be described as the ' rate of change of velocity ' as it is simply how quickly the car is increasing/decreasing in velocity. Therefore as the velocity is described as an expression of t - ti...
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Answered by Dominic E. • Maths tutor
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Find d/dx (ln(2x^3+x+8))

Use the chain rule: dy/dx = dy/du * du/dx Let y = ln(2x^3+x+8)Let u = 2x^3+x+8 dy/dx = d/dx (ln(2x^3+x+8)) = dy/du * du/dx dy/du = 1/udu/dx = 6x^2 + 1 dy/dx = 1/u * (6x^2 + 1) = (1/(2x^3+x+8)) * (6x^2 + 1) =...
Answered by • Maths tutor
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Use integration by parts to find ∫x e^(x)

So to integrate this function we would need to use a method called Integration by parts that you may have come across in your studies. This is where we separate the function into two parts to make it easier ...
Answered by • Maths tutor
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Find (dy/dx) of x^3 - x + y^3 = 6 + 2y^2 in terms of x and y

In order to differentiate this expression, we need to use implicit differentiation. An expression in the form y = f(x), where f(x) means "a function of x", is called an explicit equation and needs ...
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Differentiate y=x^2 from first principles

y=x 2 dy/dx= Lim h-->0 ( ((x+h) 2 -x 2 ) / (x+h-x) )dy/dx= Lim h-->0 ( (x 2 + 2hx + h 2 - x 2 ) / h )dy/dx= Lim h-->0 ( (h 2 + 2hx) / h )dy/dx= Lim h-->0 ( h + 2x )dy/dx= Lim h-->0 ( h ) + 2xd...
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Answered by Dipayan C. • Maths tutor
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