There's 3 rules you need to use here:
- Product Rule
- Power Rule
- Chain Rule
When taking derivatives, you always want to work outside-in. The outermost rule here is the product rule , so you'd use that.
Video, in case you need it:
The general rule is:
#d/dx[color(red)(f(x))*color(blue)(g(x))] = color(red)(d/dx[f(x)])*color(blue)(g(x)) + color(blue)(d/dx[g(x)])*color(red)(f(x))#
So:
#d/dx(5x^2sin^-1(4x)) = color(red)(d/dx(5x^2))*color(blue)(sin^-1(4x)) + color(blue)(d/dxsin^-1(4x))*color(red)(5x^2)#
That gives you two derivatives you need to evaluate.
To evaluate the first of these two derivatives, you'll need to use the power rule.
Video, in case you need it:
The general rule is:
#d/dxcolor(green)(x^a) = color(green)(ax^(a-1))#
So, what you'd have is:
#d/dxcolor(green)(5x^2) = color(green)(2(5x^(2-1)) = color(green)(10x)#
To evaluate the second of these two derivatives, you'll need to employ the chain rule.
Video, in case you need it:
The general rule is:
#d/dxcolor(red)(f(color(blue)(g(x)))) = color(red)(f'(g(x))) * color(blue)(g'(x))#
So:
#d/dxcolor(red)(sin^-1)(color(blue)(4x)) = color(red)(d/dx(sin^-1(4x))) * color(blue)(d/dx(4x))#
#color(red)(1/sqrt[1-(4x)^2] * color(blue)(4)#
#= color(orange)(4/sqrt(1-16x^2)#
Now, we just put it all together:
#d/dx(5x^2sin^-1(4x)) = color(green)(10x)*color(blue)(sin^-1(4x)) + color(blue) color(orange)(4/sqrt(1-16x^2)*color(red)(5x^2)#
Simplify, and it all boils down to:
# = color(purple)(10xsin^-1(4x) + (20x^2)/sqrt(1-16x^2)#
Hope that helps :)
