Turning point differentiation

1219269

From text book, "If the equation of a curve is known, the gradient can be determined by the process of differentiation. For example, if
y=1/2gt^2, then dy/dt = gt"

I thought that if you differentiated then you'd get -1/2t^-1x1xt, therefore -1/2t^-1xt= -1/2t^0 = 1/2?

Can someone please tell me why im completely wrong and as simply as possible explain how the answer = gt

Thanks in advance

Reply 1

Marley0103

Original post by Audi

From text book, "If the equation of a curve is known, the gradient can be determined by the process of differentiation. For example, if
y=1/2gt^2, then dy/dt = gt"

I thought that if you differentiated then you'd get -1/2t^-1x1xt, therefore -1/2t^-1xt= -1/2t^0 = 1/2?

Can someone please tell me why im completely wrong and as simply as possible explain how the answer = gt

Thanks in advance

you are differentiating with respect to t where g is a constant. (if it was say dg/dt we would use the product rule)
so, 1/2g is a multiplier of t, the power is 2.
1.multiply the power and constant

2*1/2g = g

2.then take 1 from the power of what we are differentiating with respect to, in this case t.

t^(2-1) = t^1=t

3.so our final answer is g*t = gt

Reply 2

davros

16

Original post by Audi

From text book, "If the equation of a curve is known, the gradient can be determined by the process of differentiation. For example, if
y=1/2gt^2, then dy/dt = gt"

I thought that if you differentiated then you'd get -1/2t^-1x1xt, therefore -1/2t^-1xt= -1/2t^0 = 1/2?

Can someone please tell me why im completely wrong and as simply as possible explain how the answer = gt

Thanks in advance

Are you happy that the derivative of

t^2

with respect to t is just 2t?

The (1/2)g at the front is just a constant multiplier, so when we differentiate the original expression for y we just get