Differentiate y=tanx/x?

3 Answers
Jun 20, 2018

#dy/dx=(xsec^2x-tanx)/x^2#

Explanation:

#"differentiate using the "color(blue)"quotient rule"#

#"given "y=(g(x))/(h(x))" then"#

#dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2#

#g(x)=tanxrArrg'(x)=sec^2x#

#h(x)=xrArrh(x)=1#

#dy/dx=(xsec^2x-tanx)/x^2#

Jun 20, 2018

# \frac{x-sin(x)cos(x)}{x^2cos^2(x)}#

Explanation:

You may use the quotient rule, which states that

#"d"/("d"x) \frac{f(x)}{g(x)} = \frac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}#

In you case, #f(x) = tan(x)# and #g(x) = x#

In order to compute #f'(x)#, we can use the quotient rule again, since the tangent is defined as the ratio between sine and cosine:

#"d"/("d"x) \frac{sin(x)}{cos(x)} = \frac{cos(x)cos(x)-sin(x)(-sin(x))}{cos^2(x)} = \frac{cos^2(x)+sin^2(x)}{cos^2(x)} = \frac{1}{\cos^2(x)}#

So, we have:

  • #f(x) = tan(x)#
  • #f'(x) = 1/cos^2(x)#
  • #g(x) = x#
  • #g'(x) = 1#
  • #g^2(x) = x^2#

Plug these values into the quotient rule I wrote at the beginning to get

#"d"/("d"x) \frac{tan(x)}{x} = \frac{x/cos^2(x)-tan(x)}{x^2}#

You may rewrite it as

#\frac{x/cos^2(x)-sin(x)/cos(x)}{x^2} = \frac{x/cos^2(x)-(sin(x)cos(x))/cos^2(x)}{x^2} = \frac{(x-sin(x)cos(x))/cos^2(x)}{x^2} = \frac{x-sin(x)cos(x)}{x^2cos^2(x)}#

Jun 20, 2018

#(dy)/(dx)=(xsec^2x-tanx)/x^2#

Explanation:

#y=tanx/x#

Using Quotient rule:

#color(blue)((u/v)'=(v*u'-u*v')/v^2#

Let , #u=tanx and v=x=>u'=sec^2x and v'=1#

So,

#(dy)/(dx)=((x*sec^2x)-(tanx*1))/(x)^2#

#=>(dy)/(dx)=(xsec^2x-tanx)/x^2#