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We must firstly understand that the tangent at point P is linear, and we must use y = mx +c format for the equation. To calculate the equation we need to find the unknowns m and c. To find the unknown m w...

We can use substitution for this one. Take y=ln(x) to be equal to y= 1 x ln(x)Set u=ln(x) and dv/dx=1Compute du/dx and v:du/dx=1/x and v=xUse given formula - ∫ udv/dx dx = uv - ∫ vdu/dx dx= xln(x) - ∫ x/x...

We can differentiate the terms separately:
The first term e^10x can be differentiated using the chain rule.
Let u = 10xWe can differe...

The first step is to notice that this is a standard integral in the form of 1/(x^2+a^2). In order to reach this form, we must first complete the square. Then we have 1/(x+2)^2-4+13=1/(x+2)^2+9. We can the...

Use the chain rule to differentiate the original equation: this results in 8x-3y^2*(dy/dx) + 2ln(3)3^2x=0. This can be rearranged to find dy/dx as a function of y and x: 3y^2(dy/dx)=8x+2ln(3)*3^2...