f(x) = x/2
Then the differential is lim h->0 [f(x+h) - f(x)]/h
= lim h->0 [(x+h)/2 - x/2]/h
= lim h->0 [h/2]/h
= lim h->0 [1/2] = 1/2
f(x) = x/2
Then the differential is lim h->0 [f(x+h) - f(x)]/h
= lim h->0 [(x+h)/2 - x/2]/h
= lim h->0 [h/2]/h
= lim h->0 [1/2] = 1/2
f(x) = x/2
Then the differential is lim h->0 [f(x+h) - f(x)]/h
= lim h->0 [(x+h)/2 - x/2]/h
= lim h->0 [h/2]/h
= lim h->0 [1/2] = 1/2
f(x) = x/2
Then the differential is lim h->0 [f(x+h) - f(x)]/h
= lim h->0 [(x+h)/2 - x/2]/h
= lim h->0 [h/2]/h
= lim h->0 [1/2] = 1/2
sec^2(x)
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
d/dx(x + 2) = d/dx(x) + d/dx(2) = 1 + 0 = 1
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
y' = (sec(x))^2
sec^2(x)
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
d/dx(x + 2) = d/dx(x) + d/dx(2) = 1 + 0 = 1
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
If you mean: y = x - a/x Then: y = x - ax-1 y' = 1 + ax-2 y' = 1 + a/x2 If you mean: y = (x - a)/x Then: y = 1 - ax-1 y' = ax-2 y' = a/x2
Differentiating x^2 can be accomplished by using the Power Rule. This provides that d/dx (x^2)=2x
The answer is ln(2)2x where ln(2) is the natural log of 2. The answer is NOT f(x) = x times 2 to the power(x-1). That rule applies only when the exponent is a constant.
y' = (sec(x))^2
x - 2 over 2 - x = (x - 2)/(2 - x) = -(2 - x)/(2 - x) = -1.
cos x
Yes. For example, to differentiate y = (x^2 + 1)^x, we take the natural log of both sides.ln(y) = ln((x^2 + 1)^x) Bring down the exponent. ln(y) = x ln(x^2 + 1) Differentiate both sides. dy/y = ((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Substitute in y = (x^2 + 1)^x. dy/((x^2 + 1)^x) =((2x^2)/(x^2 + 1) + ln(x^2 + 1)) dx Solve for dy/dx. dy/dx = ((x^2 + 1)^x)((2x^2)/(x^2 + 1) + ln(x^2 + 1))
2 x 2 x 3 x 3 over 2 x 2 x 2 x 2 x 3