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Differentiate implicitly with respect to x the equation x^3*y^5+3x=8y^3+1

3x²y⁵+5x³y⁴y'+3 = 24y²y'

3x²y⁵+3 = (24y²–5x³y⁴)y'

y' = (3x²y⁵

Answered by Nora G. • Maths tutor

1904 Views

f(x)=2x+c, g(x)=cx+5, fg(x)=6x+d; c and d are constants. Work out the value of d

f(g(x))=2(cx+5)+c

fg(x)=2cx+10+c

2cx=6x

2c=6

c=3

10+c=d

d=13

Answered by Imogen B. • Maths tutor

5681 Views

Make x the subject of 3y + 2x = 6y - x/2

Multiply whole equation by 2

6y + 4x = 12y - x

Take 6y from both sides

4x = 6y - x

Add x to both sides

5x = 6y

divid both sides by 5

x = 6/5(y)

Answered by Ben P. • Maths tutor

3545 Views

Multiply out: (x - 2)(x + 5)

This can be seen as the same as (x)(x + 5) + (-2)(x + 5). So you are just multiplying the first value in the first brackets with both the values in the second brackets. This is xx + x5 which equa...

Answered by Thomas S. • Maths tutor

5767 Views

What are the roots of y=x^2+5x+6 ?

Roots = where line touches x-axis. Set equation equal to 0 as y=0 at the x-axis. Now we need to solve for the x values at which y=0. Quadratic is now 0=x^2+5x+6. Factorise quadratic: 0=(x+3)(x+2). Now you...

Answered by Sam W. • Maths tutor

6637 Views