# 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

- rajLv 71 decade agoFavorite 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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- PuggyLv 71 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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- gjmb1960Lv 71 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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