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Can any one tell me the answer and the procedure to differentiate the following questions?

1)f(x) = ex2 + sin x

2)g(x) = sin(x2) sign (x square ) under root

3)h(x) = log43x2+5

4 Answers

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  • raj
    Lv 7
    1 decade ago
    Favorite Answer

    f(x)=ex^2+sinx

    f'(x)=2ex+cosx

    2.g(x)=sinx^2

    g'(x)=cosx^2*2x

    3.h(x)=log 3x^2+5

    h'(x)=1/3x^2*6x

    =2/x

    since parenthesis has not been used i have given the solution as per my understanding of the problem

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  • Puggy
    Lv 7
    1 decade ago

    1) f(x) = e^(x^2) + sinx

    Remember that the derivative of e^x is itself, e^x. Also, remember that the derivative of sin(x) is cos(x). Since we're dealing with x^2 instead of x for the first part, we have to use the chain rule and take the derivative of x^2 and multiply it to the derivative. I'll denote any usage of the chain rule with { } brackets.

    f'(x) = e^(x^2) {2x} + cosx

    Your question 2 is incomprehensible.

    3) h(x) = log[base 4](3x^2 + 5)

    The derivative of log[base 4](x) is 1/[xln4], so

    h'(x) = 1/[4ln[3x^2 + 5] ] {6x}

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  • 1 decade ago

    the procedure for all these is

    1) know the the elementary derivatives like sinx' = cosx, etc

    2) know the product rule

    thatg is all for these sums

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  • 1 decade ago

    f(x)=ex2+sinx

    f'(x)= d(ex2)/dx+ d(sinx)/dx

    = e.2x+ex2+ cosx [This has been done by applying chain rule, i.e.

    d(ex2)/dx =e.d(x2)/dx + x2.de/dx = e.2x+ex2, as d(ex)/dx = ex.

    Here x is raised to power of e.]

    g(x) = sin(x2)sin(x2) square root

    g'(x) = d(sinx2)/dx. sin(x2) square root+ sin(x2).d(sin(x2) square root)/dx

    = cos(x2).2x.sin(x2) square root+ sin(x2).[x{cos(x2)}]/sin(x2) square root

    = 3xcos(x2). sin(x2) square root

    .h(x)=log 3x^2+5

    h'(x)=1/3x^2*6x

    =2/x

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