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x = t^3 + t, y = t^2 +1, find dy/dx

dy/dx = dy/dt x dt/dx

x = t³ + t

dx/dt = 3t² +1

y = t² +1

dy/dt = 2t

dy/dx = 2t x (1 / (3t² +1) )

= 2t / (3t²

Answered by Sukhwinder K. • Maths tutor

5825 Views

Express 5cosx - 3sinx in the form Rcos(x+a).

cos(x+a)=cosxcosa-sinxsina so cosa=5 sina=3 which means tana=3/5 a=0.54 RADIANS R=sqrt(5²+3²). =sqrt(34) so sqrt(34)cos(x+0.54)

Answered by Elizabeth H. • Maths tutor

15241 Views

Differentiate sin(5x) and 3cos(x) and 3tan(5x)

sin cos -sin -cos (down is differentiate and up is integrate) so, y = sin5x dy/dx = 5cos5x y=3cox dy/dx = -3sinx y=3tan5x dy/dx = 15sec²(5x) (remember sec(x) is the reciprocal of cos(...

Answered by Abi C. • Maths tutor

4565 Views

Find where the curve 2x^2 + xy + y^2 = 14 has stationary points

d/dx (xy) = x dy/dx + y

d/dx (y^2) = 2y dy/dx [This is from the chain rule]

So, d/dx (2x^2 + xy + y^2 = 14)

=> 4x + x dy/dx + y + 2y dy/dx = 0

set dy/dx = 0 as stationary ...

Answered by Matthew H. • Maths tutor

8358 Views

Using complex numbers, derive the trigonometric identities for cos(2θ) and sin(2θ).

When dealing with complex numbers and trigonometric functions, always turn to DeMoivre's Theorem that states [cos(θ)+isin(θ)]ⁿ = [cos(nθ)+isin(nθ)]. If we set n=2, the we see a combina...

Answered by Thomas K. • Maths tutor

8164 Views