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Find the differential of y(x)=(5x*Cos(3x))^2

Firstly I would state the substitution rule, letting 5x Cos(3x)=w(x), and differentiating with respect to w(x). This gives y'(x)=2 w'(x) (w(x)).I would then demonstrate the product rule stating where u and v...
HL
Answered by Harry L. • Maths tutor
4095 Views

If x is a real number, what are the solutions to the quadratic: 4*x^2- 4*x+1 = 0

You first calculate the delta of the equation: delta = b^2-4 a c = 16 - 16 = 0 => the solution is double.x1,2 = (-b±sqrt(delta))/2/a = 4/2/4 = 1/2 which is real.so the solutions are 1/2 and 1/2.
MI
Answered by Maria I. • Maths tutor
3672 Views

The equation x^2+ kx + 8 = k has no real solutions for x. Show that k satisfies k^2 + 4k < 32.

If a quadractic equation ax^2 + bx + c = 0 has no real solutions, this means that the discriminant is less than 0, i.e. b^2-4ac&lt;0. Let's put our equation in this form: x^2 + kx + 8 = k rearranges to x^2 +...
TD
Answered by Tutor91955 D. • Maths tutor
15947 Views

Find the equation of the tangent to the curve y^3 - 4x^2 - 3xy + 25 = 0 at the point (2,-3).

This is an example of implicit differentiation where we have to consider both the y-terms as well as the x-terms. The first step is to deal with each term on the left hand side individually. This gives us 3y...
TJ
Answered by Tanmayi J. • Maths tutor
5114 Views

Differentiate with respect to x: (6x + 7)e^x

Product Rule: dy/dx (uv) = u'v + uv' Let u = 6x+7 and v = e^x. Using simple disintegartion, u' is 6 and v' is e^x. Therefore, answer is: 6e^x + (6x +7)e^x
AR
Answered by Aisha R. • Maths tutor
4269 Views