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
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.
x - 2 over 2 - x = (x - 2)/(2 - x) = -(2 - x)/(2 - x) = -1.
cos x
5 over 9 2 x 2 x 5 over 2 x 2 x 3 x 3
y = 3√x y = 3x^(1/2) y' = 3(1/2)x^(1/2 -1) y'= (3/2)x^(-1/2) y' = 3/[2x^(1/2)] y' = 3/(2√x)
y=x3+ 2x, dx/dt=5, x=2, dy/dt=? Differentiate the equation with respect to t. dy/dt=3x2*dx/dt Substitute in known values. dy/dt=3(2)2 * (5) dy/dt=60
x^2/x +2=4/x +2 (x^2/x)-(4/x)=0 (x^2-4)/x=0 x^2-4=0 (x+2)(x-2)=0 x=±2
ln2^x = xln2. let ln2 = k (constant), then the differential = k. Hence d(ln2^x)/dx = ln2
x over x is one, so the problem would be 1-2-3/2=-5/3
cosx^2 differentiates too 2(cosx)^1 x the differential of cos which is -sin so u get -2sinxcosx use the chain rule!