find the derivative by first principle f(x)= x^2-6÷3x
Answers
Answer:
Hey mate here is your answer....
We can approximate this value by taking a point somewhere near to P(x, f(x)), say Q(x + h, f(x + h)).
P
g
h
Q (x+h, f(x+h))
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Slope of the line PQ.
The value \displaystyle\frac{g}{{h}}
h
g
is an approximation to the slope of the tangent which we require.
We can also write this slope as \displaystyle\frac{{\text{change in}\ {y}}}{{\text{change in}\ {x}}}
change in x
change in y
or:
\displaystyle{m}=\frac{{\Delta{y}}}{{\Delta{x}}}m=
Δx
Δy
If we move Q closer and closer to P (that is, we let h get smaller and smaller), the line PQ will get closer and closer to the tangent at P and so the slope of PQ gets closer to the slope that we want.
P
Q
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P
Q
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Slope of the line PQ.
If we let Q go all the way to touch P (i.e. \displaystyle{h}={0}h=0), then we would have the exact slope of the tangent.
hope it's helpful for you ➿➿
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Answer:
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your answer is here !
Step-by-step explanation:
=> given that y= x²-6/3x
⇒y+бy=(x+бx)²-6/3(x+бx)
y+бy=(x²+2xбx+бx²-6)/(3x+3бx)
by subtracting y from both side
⇒бy=(x²+2xбx+бx²-6)/(3x+3бx)-y
but
=> y=x²-6/3x
⇒бy=(x²+2xбx+бx²-6)/(3x+3бx)-x²-6/3x
⇒бy={3(x²+2xбx+бx²-6)-(x²-6)(3x+бx)}/3x(3x+3бx)
=> бy/бx=(3x³+6x²бx+3xбx²-18x)-(3x³+3x²бx-18x-18бx)/3x(3x+3бx)
=> by taking the limiting value at бx⇒0
бy/бx=3x²+3xбx+18бx/9x²
⇒dy/dx=(3x²+18)/9x²
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