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dy/dx = tan³(x)sec²(x)

Integrate both sides ==> ∫dy= ∫ tan³(x)sec²(x) dx

Use the substitution u=tan(x)

And by different...

First we notice that this is a product of two functions of x, so we are going to use the product rule. Recall (uv)'(x)=u'(x)v(x)+v'(x)u(x). Let u(x)=x^(1/2) and v(x)=ln(3x). We need to find u'(x) and v'(x...

f'(x)=9x²​+4x, and f''(x)=18x+4 (derivatives) 

f'(x)=0 at x=0 or x=-4/9

when x=0 f''(x)>0 therefore a minimum value, when x=-4/9 f''(x)<0 and thus a maximum value. 

Start by rearranging the inequality - make sure the sign next to the x² term is positive to make it easier: 

2x² -5x - 12< 0

Next step is to factorise this quadratic...

Here, we're using the product rule (and the chain rule for the e^-x): y=(x²+1)(e^-x) dy/dx=(2x)(e^-x)+(x²+1)(-1)(e^-x) Then we simplify to get...