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Given that d/dx(cosx)=-sinx show that d/dx(secx)=secx(tanx)

let y=sec(x) = 1/(cos(X)) = cos(x) -1 Thus dy/dx = -1(cos(x)) -2 (-sinx) = sin(x)/(cos(x)) 2 = 1/cos(x) x sin(x)/cos(x) =sec(x)tan(x)
OD
Answered by Owain D. • Maths tutor
13634 Views

How do I use the chain rule for differentiation?

The chain rule is used when we have a function in the form f(g(x)). For example sin(x^3). [In this case, f(x) = sin(x) and g(x) = x^3] The chain rule says that the derivative of f(g(x)) is g'(x)*f'(g(x)). Fo...
TK
Answered by Tom K. • Maths tutor
5643 Views

Differentiate y^3 + 3y^2 + 5

When you differentiate, you multiply by the old power and decrease the power by 1. If the expression has a constant in it, this differentiates to 0. So the answer is 3y^(3-1) + (3x2)y^(2-1) + 0 = 3y^2 +6y
CH
Answered by Chloe H. • Maths tutor
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How would you solve (2x+16)/(x+6)(x+7) in partial fractions?

This is a simple partial fraction to solve as the denominator has already been given to you as a factorised quadratic. Because the x terms are to the dgree 1 aka x 1 we use the form (2x+16)/(x+6)(x+7) = A/(x...
KB
Answered by Katherine B. • Maths tutor
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find the value of x for when f(x)=0. f(x)=9x^(2)-4

since f(x)=0 write 9x 2 -4=0 move the x's to one side and the numbers to the other. so 9x 2 =4 then to get rid of the 9 divide both sides by 9 so x 2 =(4/9) then square root so x=sqrt(4/9) and simply by taki...
ZS
Answered by Zivanayi S. • Maths tutor
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