How do you find the derivative of #y=( (2x^3)*(e^(sinx))*(2^(cosx)) ) / (tanx-3^x)#?

Question

Isaiah Allen · Accepted Answer

#y'=(u'v-uv')/v^2# where #u'=6x^2e^sin(x)2^cos(x)+2x^3e^sin(x)cos(x)2^cos(x)+2^cos(x)ln(2)(-sin(x))#
#v=tan(x)-3^x#
#v'=1+tan^2(x)-3^xln(3)# We Need the rule #(uvw)'=u'vw+uv'w+uvw'# #(u/v)'=(u'v-uv')/v^2# #u'=6x^2e^sin(x)2^cos(x)+2x^3e^sin(x)cos(x)2^cos(x)+2^cos(x)ln(2)*(-sin(x))# #v'=1+tan^2(x)-3^xln(3)# Note that #(tan(x))'=1+tan^2(x)#

Isabella Ackerman · Answer

undefined To find the derivative of the given function y=( (2x^3)*(e^(sinx))*(2^(cosx)) ) / (tanx-3^x), we can use the quotient rule. The quotient rule states that if we have a function in the form of f(x)/g(x), then the derivative is given by (g(x)*f'(x) - f(x)*g'(x)) / (g(x))^2.

Let's denote f(x) = (2x^3)*(e^(sinx))*(2^(cosx)) and g(x) = (tanx-3^x).

Now, we need to find the derivatives of f(x) and g(x) with respect to x:

f'(x) = d/dx (2x^3*e^(sinx)*2^(cosx))
g'(x) = d/dx (tanx - 3^x)

Using the product rule for f'(x) and the derivative of tanx and 3^x for g'(x), we can compute these derivatives.

After finding f'(x) and g'(x), we can substitute these values into the quotient rule formula and simplify to get the derivative of y with respect to x.