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y=e^(2x) - x^3. Find dy/dx. (please note this is "e to the power of 2x, minus x cubed")

The rules to know are 1) when differentiating e to the power of x... be it 2x or 100x... you bring down the number in front of x, and leave the power as it is. in our case e^(2x) goes to 2e^(2x). When differ...
TM
Answered by Toby M. • Maths tutor
8312 Views

How do you find the normal to a curve at a given co-ordinate?

You first find the gradient of the tangent to the curve at this given co-ordinate by differentiating the given equation of the curve, and then, assuming the equation of the curve is in terms of x, replacing ...
SI
Answered by Srabon I. • Maths tutor
5011 Views

Find the equation of the line that is perpendicular to the line 3x+5y=7 and passes through point (-2,-3) in the form px+qy+r=0

Gradient of line 3x+5y=7: 5y=-3x+7, y=-3/5x+7/5 gradient = -3/5 Gradient of perpendicular line: 5/3 Perpendicular line with points: y+3=5/3(x+2), 3y+9=5(x+2), 3y+9=5x+10, 5x-3y+1=0
PI
Answered by Paul I. • Maths tutor
9317 Views

Differentiate y=sin(x)*x^2.

Using the chain rule, we let u = sin(x) and v = x^2. Then dy/dx = u dv/dx + v du/dx. dv/dx = 2x and du/dx = cos(x). So dy/dx = sin(x) 2x + x^2 cos(x).
LM
Answered by Lucy M. • Maths tutor
4549 Views

How to find the derivative of arctan(x)

Let y = arctan(x). Then x = tan(y). Differentiate using the chain rule and rearrange: d(x)/dx = d(tany)/dx So 1 = sec^2(y) * dy/dx dy/dx = 1/sec^2(y) But from identity sin^2(y) + cos^2(y) = 1 We can derive, ...
MR
Answered by Matthew R. • Maths tutor
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