Drawing graphs of differentiated equations (2 Viewers)

[chocolatelatte]

okay
so why is the differentiated graph of a quadratic linear? why can't it be cubic?

(now this is assuming one does not need to actually differentiate the quadratic, but only estimate the gradients relative to each other to draw a sketch of the differentiated graph. is there even a way to do it like this?)

THANKS

 

[Carrotsticks]

okay
so why is the differentiated graph of a quadratic linear? why can't it be cubic?

(now this is assuming one does not need to actually differentiate the quadratic, but only estimate the gradients relative to each other to draw a sketch of the differentiated graph. is there even a way to do it like this?)

THANKS

Think about the equation of ANY quadratic. If we differentiate it, the result will always be linear. There's no way you can differentiate a quadratic and then somehow wind up with a cubic.

 

[chocolatelatte]

ok maybe I'm just dumb
but x^2 is obviously linear if you differentiate it.
but i want to draw this graph without differentiating it and only going off the gradients of x^2
and x^2 increases when x>0
and decreases when x<0
and if i drew a cubic , say x^3, wouldn't it satisfy this?
a linear implies that the gradient is increasing at a constant rate and idk but i can't seem to grasp that the gradients of a quadratic increase this way D=

 

[chocolatelatte]

ALSO,
if i was to draw the differentiated graph of an exponential , the differentiated graph too would be an exponential, but can someone please explain to me why this differentiated graph (which is an exponential) also has the y-intercept 1?

 

[QZP]

ok maybe I'm just dumb
but x^2 is obviously linear if you differentiate it.
but i want to draw this graph without differentiating it and only going off the gradients of x^2
and x^2 increases when x>0
and decreases when x<0
and if i drew a cubic , say x^3, wouldn't it satisfy this?
a linear implies that the gradient is increasing at a constant rate and idk but i can't seem to grasp that the gradients of a quadratic increase this way D=

I know what you mean. This is basically what you're saying:
Consider y = x^4. Its derivative is a cubic. Now, since y = x^4 and y = x^2 "look alike", there is no visual way to confirm the derivative of y = x^2 is linear.

You'd have to differentiate it algebraically.

 

[QZP]

ALSO,
if i was to draw the differentiated graph of an exponential , the differentiated graph too would be an exponential, but can someone please explain to me why this differentiated graph (which is an exponential) also has the y-intercept 1?

Not true.

 

[chocolatelatte]

OHHH ok THANK YOU!!!!! i just couldn't seem to understand it considering these exercises were all graphing by visual differentiation
ohhhh well the answers that said it intercepted at 1 must be wrong then. that makes a lot more sense , lol