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Given that y = 8x + 2x^-1, find the 2 values for x for which dy/dx = 0

First differentiate y with respect to x, which gives you dy/dx = 8 - 2x^-2. This needs to equal zero so equate to zero. 8-2x^-2 = 0. You can then bring the 2x^-2 to the other side giving 2x^-2=8. Dividing...

RB
Answered by Rosemary B. Maths tutor
3452 Views

Differentiate y = 4exp(6x) + cos(x) + 6x

dy/dx = 24e6x - sin(x) + 6

BC
Answered by Benn C. Maths tutor
3922 Views

Solve the simultaneous equations: 3x+2y = 11, 2x-5y=20

Firstly, solving simultaneous equations finds the point(s) where two lines on a graph meet. They can either be two linear equations (finding one point of intersection), a linear and a quadratic (can have ...

OC
Answered by Olivia C. Maths tutor
9984 Views

Express 2 cos x – sin x in the form Rcos( x + a ), where R and a are constants, R > 0 and a is between 0 and 90 ° Give the exact value of R and give the value of to 2 decimal places.

2cosx - sinx = Rcos(x+a) = Rcos(x)cos(a)-Rsin(x)sin(a)

Implying

2cos(x)=Rcos(x)cos(a)

Rcosa = 2

Similarly: Rsin(a) = 1

Therefore tan(a) = 1/2

Meaning a=26.57 Degr...

TO
Answered by Thomas O. Maths tutor
30676 Views

A particle is moving along a straight line. The fixed point O lies on this line. The displacement of the particle from O at time t seconds is s metres where s = 2t3 – 12t2 + 7t(a) Find an expression for the velocity, v m/s, of the particle at time t.

To find velocity you differentiate displacement with respect to time

Therefore ds/dt = v = (3x2)t2-(12x2)t2+7

Implying v = 6t2-24t+7

TO
Answered by Thomas O. Maths tutor
23207 Views

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