Differentiating the given function #f(x)# is determined by using
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the product rule differentiation.
Let #color(blue)(h(x)=tanx^2 and g(x)=sec sqrt (x)#
#f(x)=color(blue)(h(x))xxcolor(blue)(g(x))#
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#color(red)((df(x))/dx = (dh(x))/dxxxcolor(blue)(g(x)) + (dg(x))/dxxxcolor(blue)(h(x))#
Let us differentiate #color(blue)(h(x))#
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#color(blue)(h(x))# is a composite of two functions #color(green)(u(x)=tanx and v(x)=x^2)# ,
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so #color(blue)(h(x))# is differentiated by applying chain rule.
Knowing the differentiation of #tanx#: #" "##color(green)(u'(x)=(dtanx)/dx=sec^2x#
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Applying chain rule in differentiating #color(blue)(h(x))#
#color(red)((dh(x))/dx=(dtanx^2)/dx=u'(v(x))xxv'(x)=sec^2(v(x)) xx 2x=2xsec^2x^2#
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Let us differentiate #color(blue)(g(x))#
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#color(blue)(g(x))# is a composite of two functions #color(green)(u(x)=secx and v(x)=sqrtx)#
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so #color(blue)(g(x))# is differentiated by applying chain rule.
Knowing the differentiation of #secx#:
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#color(green)((dsecx)/dx=tanxsecx#
The derivative of #color(green)(sqrtx#:#" "# #color(green)((dsqrtx)/dx=1/(2sqrtx))#
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Applying chain rule in differentiating #color(blue)(g(x))#
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#color(red)((dg(x))/dx)=(d(u(v(x))))/dx=u'(v(x)) xx v'(x)=tan(v(x))sec(v(x))xx1/(2sqrtx)= tan(sqrtx)sec(sqrtx)xx1/(2sqrtx)#
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#color(red)((df(x))/dx = (dh(x))/dxxxcolor(blue)(g(x)) + (dg(x))/dxxxcolor(blue)(h(x))#
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#color(red)((df(x))/dx = 2xsec^2x^2xxcolor(blue)(sec sqrt (x)) + tan(sqrtx)sec(sqrtx)xx1/(2sqrtx)xxcolor(blue)(tanx^2)#
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#(d(f(x)))/dx=sec(sqrtx)secx^2(2xsecx^2+(tansqrtxsinx^2)/(2sqrtx))#
