J's answer to Candice's Junior College 1 H2 Maths Singapore question.
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This question is very similar to the differentiation of x^x
Firstly, let y = x^(sec2x)
ln y = ln x^(sec2x)
ln y = sec2x ln x
Next, differentiate implicitly, for both sides, with respect to x.
We'll need the product rule and the knowledge that d/dx sec f(x) = f'(x) sec f(x) tan f(x)
So, differentiating both sides with respect to x,
1/y dy/dx = 2 sec2x tan2x lnx + sec2x (1/x)
dy/dx = y (2 sec2x tan2x lnx + sec2x / x)
dy/dx = x^(sec2x) (2 sec2x tan2x lnx + sec2x / x)
dy/dx = x^(sec2x) sec2x (2 tan2x lnx + 1/x)
or x^(sec2x - 1) sec2x (2x tan2x lnx + 1)
Expand or factorise the final answer accordingly
Firstly, let y = x^(sec2x)
ln y = ln x^(sec2x)
ln y = sec2x ln x
Next, differentiate implicitly, for both sides, with respect to x.
We'll need the product rule and the knowledge that d/dx sec f(x) = f'(x) sec f(x) tan f(x)
So, differentiating both sides with respect to x,
1/y dy/dx = 2 sec2x tan2x lnx + sec2x (1/x)
dy/dx = y (2 sec2x tan2x lnx + sec2x / x)
dy/dx = x^(sec2x) (2 sec2x tan2x lnx + sec2x / x)
dy/dx = x^(sec2x) sec2x (2 tan2x lnx + 1/x)
or x^(sec2x - 1) sec2x (2x tan2x lnx + 1)
Expand or factorise the final answer accordingly
Date Posted:
3 years ago
For enrichment : if given x^x,
let y = x^x
ln y = ln(x^x)
ln y = x lnx
Differentiate both sides with respect to x,
1/y dy/dx = lnx + x(1/x)
1/y dy/dx = lnx + 1
dy/dx = ylnx + y
dy/dx = x^x lnx + x^x
let y = x^x
ln y = ln(x^x)
ln y = x lnx
Differentiate both sides with respect to x,
1/y dy/dx = lnx + x(1/x)
1/y dy/dx = lnx + 1
dy/dx = ylnx + y
dy/dx = x^x lnx + x^x
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