Find the equations of the normals to the curves at the given point
y=(x+1)^2 at x=0
y=4/x at x=1
answers in form ax+by+c=0
how do i do this
also can some1 check if this is right
find the equations of the tangents to the curves at the given points
y=5x^2+1 at x=2 y=50?
y=3x^24x at x=3 y=14??
need help on dy/dx


(Original post by mr tim)
For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate , and then substitute your values for x to get the gradient.
Similar thing for except you differentiate instead. 
(Original post by mr tim)
For the first part is not right, apart from x=2. To calculated the gradient of the tangent, you differentiate , and then substitute your values for x to get the gradient.
Similar thing for except you differentiate instead.
x=2 gradient=4
x=6 gradient=12
x=1 gradient=1? 
(Original post by jetskiwavedunkno)
ah wait for the first ones is it
x=2 gradient=4
x=6 gradient=12
x=1 gradient=1? 
(Original post by jetskiwavedunkno)
ah wait for the first ones is it
x=2 gradient=4
x=6 gradient=12
x=1 gradient=1? 
(Original post by Gemini92)
The first two are right, but the last one is wrong. 
(Original post by raheem94)
Your calculated gradient at x=1 is wrong, others are correct. 
(Original post by jetskiwavedunkno)
is the gradient 2? nd how do i do the x^3 ones plz
For the x^3 expression, first differentiate it, than substitute different values of x to find the gradient at different points. 
(Original post by jetskiwavedunkno)
is the gradient 2? nd how do i do the x^3 ones plz
Eg. differentiating x^6 =6x^5.
Yeah, 2's correct. 2(1)=2.
Try using the rule I stated above to differentiate x^3. Actually write what you get in terms of x first so we can see what you're doing. 
(Original post by jetskiwavedunkno)
x=1 gradient= 2??? how do i do the x^3 ones 
For the first one, you need to start by differentiating x^2.
Differentiating it gives you "dy/dy", also known as the gradient function.
The method of differentiating, is "Multiply by the current power, and then decrease the power by 1."
So to differentiate x^2, you multiply by 2, and decrease the power fromk 2 to 1; hence the result is 2x.
Youy would then need to substitute the value of x in to find the gradient at that point.
For x^3, it is a similar process. Start by multiplying by the power, so multiply by 3, and then reduce the power from 3 to 2. Hence the result is 3x^2. Again sub in values of x to find the gradient. 
(Original post by raheem94)
Yes the gradient is 2.
For the x^3 expression, first differentiate it, than substitute different values of x to find the gradient at different points.
x=3 gradient=27
x=5 gradient=75
x=3 gradient=27
???? 
(Original post by Contrad!ction.)
To differentiate, you'll multiply x by its power, and then subtract 1 from the power.
Eg. differentiating x^6 =6x^5.
Yeah, 2's correct. 2(1)=2.
Try using the rule I stated above to differentiate x^3. Actually write what you get in terms of x first so we can see what you're doing.
x=3 gradient=27
x=5 gradient=75
x=3 gradient=27
???? 
(Original post by Gemini92)
That's right. For the x^3 ones, you have to differentiate x^3 and then substitute each x value into the derivative to get the gradient.
x=3 gradient=27
x=5 gradient=75
x=3 gradient=27
???? 
(Original post by jordans)
For the first one, you need to start by differentiating x^2.
Differentiating it gives you "dy/dy", also known as the gradient function.
The method of differentiating, is "Multiply by the current power, and then decrease the power by 1."
So to differentiate x^2, you multiply by 2, and decrease the power fromk 2 to 1; hence the result is 2x.
Youy would then need to substitute the value of x in to find the gradient at that point.
For x^3, it is a similar process. Start by multiplying by the power, so multiply by 3, and then reduce the power from 3 to 2. Hence the result is 3x^2. Again sub in values of x to find the gradient. 
No.. differentiating x^3 gives 3x^2.. sub in x = 3 and you get 3(3)^2 = 27.

(Original post by jetskiwavedunkno)
is it for x^3
x=3 gradient=18
x=5 gradient=50
x=3 gradient=18
????
Differentiating x^3 gives 3x^2. 
(Original post by jetskiwavedunkno)
ahhh crap wait so differentiating 3x is 3x^2 and not 2x^2??
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