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How do I differentiate a function of x and y with respect to x?

To differentiate a function of x and y, you must differentiate x as you would ordinarily, and then differentiate y as you would normally, but multiply the differentiated term by dy/dx. For terms with x and y...
HM
Answered by Hannah M. • Maths tutor
31382 Views

Find the derivative of f(x)=exp((tanx)^(1/2))

We use the chain rule. Let u(x)=exp(x), v(x)=x 1/2 , w(x) = tan(x). Then f(x) = u(v(w(x))). So by the chain rule, f'(x) = u'(v(w(x)))*(v(w(x)))'. u'(x) = exp(x). By the chain rule, (v(w(x)))' = v'(w(x)) w'(x...
LD
Answered by Luke D. • Maths tutor
4449 Views

Solve the equation 3^(2x+1)=1000

We can use logarithms with base 3 to solve this equation.2x+1=log 3 10002x=log 3 (1000)-1x=0.5(log 3 (1000)-1)x=2.64
SD
Answered by Samuel D. • Maths tutor
5536 Views

Integrate 6/(e^x+2) between 1 and 0

By Trapezium Rule (with 5 segments)x 0 0.2 0.4 0.6 0.8 1y 2 1.8625 1.71830 1.56981 1.41994 1.27165≈1⁄2*(0.2) [2+1.27165+ 2( 1.8625 + 1.71830 + 1.56981 + 1.41994)]≈1⁄10 *16.41283 = 1.641By Substitution/ Parti...
CH
Answered by Christopher H. • Maths tutor
4869 Views

Binomially expand the equation (2+kx)^-3

(2+kx) -3 = (2 -3 )(1+kx/2) -3 = (2 -3 )(1+(-3)(kx/2) + [(-3)(-4)]/2! (kx/2) 2 + [(-3)(-4)(-5)]/3! (kx/2) 3 +... ) = 1⁄8 [1 -(3kx/2) + (12⁄2 k 2 x 2 /4) + (60⁄6 k 3 x 3 /8) + ...] = 1⁄8 [1 - (3⁄2 kx) + ( 3⁄2...
CH
Answered by Christopher H. • Maths tutor
9799 Views