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Find the gradient at the point (0, ln 2) on the curve with equation e^2y = 5 − e^−x . [4]

Figure out what skills are being tested: implicit differentiation and exponentials and logarithms.e^2y= 5 - e^-x2e^2y(dy/dx) = e^-x(dy/dx) = e^-x/ 2e

Answered by Theodore C. • Maths tutor

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Find dy/dx when x+2y+3y^2= 2x^2+1

To solve this question we can use implicit differentiation. We can write:d/dx(x+2y+3y^2)=d/dx(2x^2+1).When differentiating something in terms of y with respect to x we can use the chain rule, this allows ...

Answered by Adam G. • Maths tutor

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Solve D/dx (ln ( 1/cos(x) + tan (x) )

Solve D/dx (ln ( 1/cos(x) + tan (x) ) As always, we approach with a substitution method that we would normally use for differentiating ln (x). So, we try differentiate ln (t) with t= 1/cos(x) + tan(x)So w...

Answered by Amera D. • Maths tutor

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let line L have the equation 4y -3x =10, and line M passes through the points (5,-1) and (-1,8), find out if they are perpendicular, parallel, or neither

L:4y=3x+10 -> y=3/4 x + 10/3. M: y= mx+ c such that it passes through the points in question, then m=(-1-8)/(5-(-1))=-3/2. As the gradient of L is not a negative reciprocal of the gradient of M, nor is...

Answered by Illya N. • Maths tutor

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