Differentiating the given function f(x)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)h(x)=tanx2andg(x)=sec√x
f(x)=color(blue)(h(x))xxcolor(blue)(g(x))f(x)=h(x)×g(x)
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color(red)((df(x))/dx = (dh(x))/dxxxcolor(blue)(g(x)) + (dg(x))/dxxxcolor(blue)(h(x))df(x)dx=dh(x)dx×g(x)+dg(x)dx×h(x)
Let us differentiate color(blue)(h(x))h(x)
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color(blue)(h(x))h(x) is a composite of two functions color(green)(u(x)=tanx and v(x)=x^2)u(x)=tanxandv(x)=x2 ,
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so color(blue)(h(x))h(x) is differentiated by applying chain rule.
Knowing the differentiation of tanxtanx: " " 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))
