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We will solve this question with the knowledge that dy/dx = u.(dv/dx) + v.(du/dx), where y=u.v We have y=e^x(3x+1)^2. First, we want to find u & v. By splitting the function, we have that u=e^x and v=...

We have 2tan(th) / (1 + tan^2(th)) = sin(2th)

We know that tan(A) = sin(A) / cos(A), and 1 + tan^2(A) = sec^2(A)

Therefore => (2sin(th) / cos(th)) / sec^2(th)

=> 2sin(th)*cos^2(...

We have 3^(2x+1) = 4^100

=> log(3^(2x+1)) = log(4^100)

=> (2x+1)log(3) = 100log(4)

The first step is to differentiate both sides of this equation with respect to x - we will then be able to solve for dy/dx. Differentiating the right side of the equation gives d/dx(17)=0. We’ll different...

We want to rearrange the expression to the form (1+y)^n so we can use the general result: (1+y)^n=1+ny+[n(n-1)/2]y^2+[n(n-1)(n-2)/3!]y^3+... 1/(2+5x)^3 = (2+5x)^-3 = [2(1+5x/2)]^-3 = (2^-3)(1+5x/2)^-3 usi...