2016年12月1日 · f' (x)=-csc^2x differentiate f (x) using the color (blue)"quotient rule" color (orange)"Reminder" "If " f (x)= (g (x))/ (h (x)) " then" color (red) (bar (ul (|color ...

2015年8月17日 · dy/dx = (y [cos (xy) + sin x])/ (cos x - x cos (xy)) Always think of your chain, product and quotient rules when differentiating y cos x = 1 + sin (xy) Diff wrt x: dy ...

2016年8月2日 · Let y = cosu and u = lnx. Then y' = -sinu and u' = 1/x f'(x) = -sinu xx 1/x = -sin(lnx) xx 1/x = -(sin(lnx))/x Hopefully this helps!

2018年1月22日 · (df)/ (dg)=sqrt ( (2-x^2)/ (1-x^2)) Let f (x)=cos^ (-1) (2x^2-1) and g (x)=sin^ (-1) (1-x^2) what we are seeking is (df)/ (dg), which is equal to ( (df)/ (dx))/ ( (dg ...

2015年9月13日 · 2) Then multiply that by the derivative of the inner function. In cos (x^2-4x), the outer function is cos x and the inner function is x^2 - 4x.

2016年11月12日 · See below. You could use the product rule and the chain rule, but I find it more satisfying to rewrite the function before differentiating. From trigonometry: sin (2theta) = 2sin (theta)cos (theta) Therefore sin (theta)cos (theta) = 1/2sin (2theta) So, f (x) = cos (3x)sin (3x) = 1/2sin (6x) Therefore, f' (x) = 1/2cos (6x) * d/dx (6x) " " (chain rule) = 3cos (6x)

2017年6月16日 · d/ (dx)cos^10x=-10sinxcos^9x To differentiate y=cos^10x, le us use chain rule. Let y= (g (x))^10 and g (x)=cosx Now we have (dy)/ (dg)=10 (g (x))^9 and (dg)/ (dx ...

2018年2月1日 · cos2x = [cosx]2 and so need to use the chain rule to differentiate this, = 2cosx, times the derivative of cosx which is - sinx, so d dx cos2x = − 2sinxcosx. hope this helped.

2016年2月15日 · -sin (tan (x))sec^2 (x) We can use the chain rule to create a rule for differentiating a cosine function: Since d/dx (cos (x))=-sin (x), we know that d/dx (cos (u ...

2016年10月3日 · Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos (x) and y=tan (x)